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Fynjy0 [20]
3 years ago
5

The inductor in the RLC tuning circuit of an AM radio has a value of 250 mHmH . You may want to review (Pages 857 - 860) . Part

A What should be the value of the variable capacitor in the circuit to tune the radio to 731 kHz kHz
Physics
1 answer:
ANEK [815]3 years ago
6 0

Answer:

The value of variable capacitor is 1.89 \times 10^{-13} F

Explanation:

Given :

Inductance L = 250 \times 10^{-3} H

Frequency f = 731 \times 10^{3} Hz

According to the cutoff frequency,

   f = \frac{1}{2\pi \sqrt{LC} }

Now we find the value of capacitance,

  C = \frac{1}{4\pi ^{2} f^{2}  L }

  C = \frac{1}{4\times 9.85 \times (731 \times 10^{3} )^{2} \times 250 \times 10^{-3}  }

  C = 1.89 \times 10^{-13} F

Therefore, the value of variable capacitor is 1.89 \times 10^{-13} F

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How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

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next we find Initial velocity

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now we find the number of moles

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N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

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Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

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