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2 years ago

What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit

Physics

1 answer:

Tresset [83]2 years ago

4 0

The highest frequency sound to which the machine can be adjusted is :

4179.33 Hz

<u>Given data :</u>

Pressure = 10 Pa

Speed of sound = 344 m/s

Displacement altitude = 10⁻⁶ m

<h3>Determine the highest frequency sound ( f ) </h3>

applying the formula below

Pmax = $B(\frac{2\pi f}{v}) A$ --- ( 1 )

Therefore :

f = ( Pmax * V ) / $2\pi \beta A$

= ( 10 * 344 ) / 2 $\pi$ * 1.31 * 10⁵ * 10⁻⁶

= 4179.33 Hz

Hence we can conclude that The highest frequency sound to which the machine can be adjusted is : 4179.33 Hz .

Learn more about Frequency : brainly.com/question/25650657

<u><em>Attached below is the missing part of the question </em></u>

<em>A loud factory machine produces sound having a displacement amplitude in air of 1.00 μm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.31×105 Pa. The speed of sound in air is 344 m/s. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?</em>

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stealth61 [152]

50 μm = 5 × 10⁻⁵ m in diameter

<h3>Steps</h3>

1μm = 1 x $10^{-6}$ m

<h3>Given</h3>

1μm = 1 x $10^{-6}$ m

50μm = 50 x $10^{-6}$ m

50μm = 50 x 10 x $10^{-6}$

50μm = 5 x $10^{-5}$ m

<h3>Conclusion</h3>

A human hair is approximately 50 μm = 5 × 10⁻⁵ m in diameter

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between 17 and 181 micrometers

The typical diameter of human hair can range from 17 micrometers to 181 micrometers, according to research. Using 50 micrometers as an average figure for its diameter, this translates to 50,000 nanometers.

<h3>What is a human hair's diameter in inches?</h3>

The finest hair is flaxen, with a diameter ranging from 1/1500 to 1/500 of an inch. the coarsest hair is black, measuring between 1/450 and 1/140 of an inch.

learn more about human hair diameter here

<u>brainly.com/question/13147893</u>

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1 year ago

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$

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77julia77 [94]

Answer:

a) dB / dA = 2 ,

b) Network B Network A

2 1

4 2

6 3

Explanation:

a) The expression for grating diffraction is

d sin θ = m λ

where d the distance between two slits, λ the wavelength and m an integer that represents the diffraction range

In this exercise we are told that the two spectra are in the same position, let's write the expression for each network

Network A

m = 1

sin θ = 1 λ / dA

Network B

m = 2

sin θ = 2 λ / dB

they ask us for the relationship between the distances, we match the equations

λ / dA) = 2 λ / dB

dB / dA = 2

b) let's write the equation of the networks

sin θ = m_A λ / dA

sin θ = m_B λ / dB

we equalize

m_A λ/ dA = m_B λ / dB

we use that

dB / dA = 2

m_A 2 = m_B

therefore the overlapping orders are

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3 years ago

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FromTheMoon [43]

Answer:

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Answer:

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$F=\dfrac{mv^2}{r}$