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Rzqust [24]
3 years ago
14

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

se also that a sinusoidal potential difference with a maximum value and a frequency of =60 Hz is applied across the plates. Find , the maximum value of the induced magnetic field that occurs at r = R. Express your answers in terms of variables first and then plug in numbers afterword.
Physics
1 answer:
seropon [69]3 years ago
7 0

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

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LekaFEV [45]

Answer:

Height is 11.25m

Explanation:

<u>Given the following data;</u>

Initial velocity, u = 0

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To find the height, we would use the third equation of motion;

V^{2} = U^{2} + 2aS

Where;

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V^{2} = U^{2} + 2aS

<em>Making S the subject, we have;</em>

S = \frac {V^{2} - U^{2}}{2a}

But a = g = 10m/s²

<em>Substituting into the equation, we have;</em>

S = \frac {15^{2} - 0^{2}}{2*10}

S = \frac {225 - 0}{20}

S = \frac {225}{20}

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2 years ago
It takes Seymour, a slimy slug, 20 minutes to travel from his favorite bush to the local trash can (a distance of 30 meters), ho
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The human ear canal is about 2.9 cm long and can be regarded as a tube open at one end and closed at the eardrum. What is the fu
solniwko [45]

The frequency of the human ear canal is 2.92 kHz.

Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is  

Wavelength=4*Length of the ear canal

As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

wavelength = 4*2.9*10^{-2} =0.116

Then the frequency is determined as

f=c/λ=339/0.116=2922 Hz=2.92 kHz.

So, the frequency of the human ear canal is 2.92 kHz.

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If the particles of an object have more kinetic energy than the particles of an object, what must be true?
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(b) Can the speed of a rocket exceed the exhaust speed of the fuel? Explain.
muminat

<u>Yes. The speed of a rocket can exceed the exhaust speed of the fuel.</u>

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To know more about conservation of momentum, refer:

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