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3 years ago

What’s the velocity of a sound wave traveling through air at a temperature of 18°C (64.4°F)?

Physics

2 answers:

GREYUIT [131]3 years ago

7 0

The velocity of a sound wave traveling through air at a temperature of 18°C (64.4°F) is 343m/s.

Explanation :

At normal atmosphere and pressure, the velocity of sound wave at increase in temperature is as follows

$\mathrm{V}=331 \mathrm{m} / \mathrm{s}+(0.6 \mathrm{m} / \mathrm{s} / \mathrm{c})^{*} \mathrm{T}$

where T is the temperature in degrees Celsius.

Using this equation, calculate the velocity of a sound wave at 20 degrees Celsius , from the following equations,

$\mathrm{V}=331 \mathrm{m} / \mathrm{s}+(0.6 \mathrm{m} / \mathrm{s} / \mathrm{c})^{*} \mathrm{T}$

$\mathrm{v}=331+(0.6)^{*} 20$

$\mathrm{v}=331+12$

V = 343 m/s

Anna11 [10]3 years ago

4 0

Answer:

342 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331.4 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 18:

v ≈ 331.4 + 0.6 (18)

v ≈ 342.2

The velocity is approximately 342 m/s.

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

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Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

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First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

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$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

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