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Nataliya [291]
3 years ago
12

A capacitor is connected to an AC generator. As the generator's frequency is increased, what happens to the current in the capac

itor
Physics
1 answer:
Licemer1 [7]3 years ago
8 0

Answer:

The current increases.

Explanation:

A capacitor can be defined as an electronic component used in electrical circuits to store charge temporarily.

A capacitor is connected to an AC generator. As the generator's frequency is increased, the current in the capacitor increases as well.

This ultimately implies that, when a capacitor is connected to an AC generator, the frequency of the circuit is directly proportional to the amount of current flowing through it.

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Wavelength - the distance from one wave crest or trough to another wave crest or trough. Amplitude - the distance from the median point or "middle" of the wave straight up to a crest (a maximum) or straight down to a trough (or minimum), which is the peak amplitude; or the distance from a trough straight up to a crest, or a crest straight down to a trough, called peak-to-peak amplitude.
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Using Planck’s constant as h = 6.63 E-34 J*s, what is the wavelength of a proton with a speed of 5.00 E6 m/s? The mass of a prot
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De Broglie's identity gives the relationship between the momentum and the wavelength of a particle:
p=mv= \frac{h}{\lambda}
where
p is the particle momentum
m is its mass
v its velocity
h is the Planck constant
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By re-arranging the equation, we get
\lambda=  \frac{h}{mv}
and by using the data about the proton, given in the text, we can find the proton's wavelength:
\lambda= \frac{h}{mv} = \frac{6.63 \cdot 10^{-34} Js}{(1.66 \cdot 10^{-27} kg)(5.00 \cdot 10^6 m/s)} =7.99 \cdot 10^{-14} m
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3 years ago
A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the cont
xenn [34]

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W =  (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

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3 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

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Risk of not being able to reduce their weight
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