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Setler [38]
3 years ago
14

At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at 1.5 atm pressure an

d -0.77°C? Take the diameter of an oxygen molecule to be 8.4 x 10^-8 cm and the speed of sound to be 330 m/s.
Physics
1 answer:
quester [9]3 years ago
5 0

Answer:

\vartheta = 41.31 GHz

Given:

Pressure, P_{O} = 1.5 atm = 1.5\times 10^{5} Pa

Temperature, T = - 0.77^{\circ} = 273 + (- 0.77) = 272.23 K

Diameter of oxygen molecule, d_{O} = 8.4\times 10^{- 8} cm = 8.4\times 10^{-10} m

Speed of sound, v = 330 m/s

k_{B} = 1.38\times 10^{- 23} J/s

Solution:

Mean free path of oxygen is given by:

L_{O} = \frac{k_{B}T}{\sqrt{2}P_{O}\pi d_{O}^{2}}

Now, substituting the given values in the above formula:

L_{O} = \frac{1.38\times 10^{- 23}\times 272.23}{\sqrt{2}\times 1.5\times 10^{5}\pi\times (8.4\times 10^{- 10})^{2}}

L_{O} = 7.99\times 10^{-8} m

Now, the frequency, \vartheta is given by:

\vartheta = \frac{c}{\lambda}  

Since, mean free path = wavelength = 7.99\times 10^{-9} m      

Therefore,

\vartheta = \frac{v}{L_{O}}  

\vartheta = \frac{330}{7.99\times 10^{-8}}  

\vartheta = 4.131\times 10^{10} Hz = 41.31 GHz

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Answer:

0.000003782 m

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D = Diameter of beam = 1 cm

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The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

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d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
3 years ago
What is the distance covered in 20 minutes by a train traveling 500m/m?
MrMuchimi

Answer:

answer is 10km

Explanation:

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3 years ago
when the mass of an object is increased it would decrease its acceleration in the force is left alone
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Answer:

See the explanation below.

Explanation:

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∑F = m*a

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m = mass [kg]

a = acceleration [m/s²]

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a=\frac{F}{m}

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3 years ago
An RLC circuit is used in a radio to tune into the radio lagos fm Station broadcasting at 93.5Hz. The resistance is 15ohms and t
bazaltina [42]

The characteristics of the RLC circuit allow to find the result for the capacitance at a resonance of 93.5 Hz is:

  • Capacitance is C = 1.8 10⁻⁶ F

A series RLC circuit reaches the maximum signal for a specific frequency, called the resonance frequency, this value depends on the impedance of the circuit.

            Z^2 = R^2 + ( wL - \frac{1}{wC} )^2  

Where Z is the impedance of the circuit, R the resistance, L the inductance, C the capacitance and w the angular velocity. The negative sign is due to the fact that the current in the capacitor and the inductor are out of phase.

In the case of resonance, the impedance term completes the circuit as a resistive system.

           wL - \frac{1}{wC} = 0  \\w^2 = \frac{1}{LC}  

           

Indicate that the inductance L = 1.6 H and the frequency f = 93.5 Hz.

Angular velocity and frequency are related.

         

         w = 2π f

           

Let's  substitute.

          C = \frac{1}{L ( 2 \pi f)^2 }  

 

Let's calculate.

         C = \frac{1}{1.6 \ ( 2\pi \ 93.5)^2}  

         C = 1.8 10⁻⁶ F

In conclusion with the characteristics of the RLC circuits we can find the result for the capacitance at a 93.5 Hz resonance is:

  • Capacitance is C = 1.8 10⁻⁶ F

Learn more about serial RLC circuits here: brainly.com/question/15595203

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