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Annette [7]
2 years ago
8

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?
Physics
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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kvasek [131]

Answer:

The maximum electric field strength = 0.01 V/m

Explanation:

Given

ΔV(max) = 4.00 mV = 0.004 V

d = 0.400 m

f = 1.00 Hz

Maximum electric field = (maximum potential)/(length)

Maximum electric field = E(max)

Maximum potential = 4.00 mV = 0.004 V

Length = 0.400 m

E(max) = (0.004/0.4) = 0.01 V/m

Hope this Helps!!!

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2 years ago
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3 0
3 years ago
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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

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

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\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}

And the angular frequency can be written as

\omega = 2\pi f

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

PART B) The new amplitude A' that is half ot the wavelength of the wave is

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)})

P = 0.058W

8 0
2 years ago
True or False: For a longitudinal wave, the wavelength is the distance between compressions.
Schach [20]

Answer:

false....

Explanation:

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2 years ago
Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o
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Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

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

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

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

8 0
2 years ago
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