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2 years ago

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An MRI machine needs to detect signals that oscillate at very high frequencies. It does so with an LC circuit containing a 15mH

coil. To what value should the capacitance be set to detect a 450 MHz signal?

1 answer:

vfiekz [6]2 years ago

0 0

Answer:

The capacitance is $C = 3.2 9 *10^{-16} \ F$

Explanation:

From the question we are told that

The induction of the LC circuit is $L = 15 mH = 15 *10^{-3} \ H$

The frequency is $w = 450 \ MHz = 450 *10^{6} \ Hz$

The natural frequency is mathematically represented as

$w = \frac{1}{\sqrt{LC} }$

Where C is the capacitance So

=> $C = \frac{1}{L * w^2}$

substituting values

$C = \frac{1}{15 *10^{-3} * [450 *10^{6}]^2}$

$C = 3.2 9 *10^{-16} \ F$

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Given

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The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

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$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

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Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

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$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

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Answer:

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Explanation:

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r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

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Now from energy balance

Q = ρ×A×v

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⇒ x = 455.53-70

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∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

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