What is the speed of sound at a temperature

of 56°c?

a: 316 meters/second

b: 331 meters/second

c: 365 meters/second

d: 373 meters/second

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of 56°c?

a: 316 meters/second

b: 331 meters/second

c: 365 meters/second

d: 373 meters/second

36,67 degrees Celsius

Explanation:

The simplest way to approach this problem, given the information provided, is to simply start with the speed difference.

Goal: 353 m/s

Start: 343 m/s (at 20 degrees Celsius).

Difference: 10 m/s

Variation rate: 0.60 m/s/d (d = degree)

[tex]d = \frac{10 m/s}{0.60 m/s/d} = 16,67 d[/tex]

So, 16,67 degrees more than the starting point.

The temperature will then be 36.67 degrees Celsius, when the sound travels at the speed of 353 m/s.

The correct answer is Option C.

Explanation:

Speed of sound is directly related to the temperature.

The relation between speed of sound and temperature is given by the equation:

[tex]\text{Speed of sound}=331.5+0.60T(^oC)[/tex]

where, T = absolute temperature

Putting T = 56°C in above equation, we get:

[tex]\text{Speed of sound}=331.5+0.60\times 56=365.1m/s[/tex]

Hence, the correct option is Option C.

The correct answer is Option C.

Explanation:

Speed of sound is directly related to the temperature.

The relation between speed of sound and temperature is given by the equation:

[tex]\text{Speed of sound}=331.5+0.60T(^oC)[/tex]

where, T = absolute temperature

Putting T = 56°C in above equation, we get:

[tex]\text{Speed of sound}=331.5+0.60\times 56=365.1m/s[/tex]

Hence, the correct option is Option C.

0.69444 m, 0.08152 m, 0.32407 m, 0.03804 m

Explanation:

v = Velocity of sound

f = Frequency

Length of vocal tract is given by

[tex]L=\dfrac{v}{4f}[/tex]

At f = 270 Hz v = 750 m/s

[tex]L=\dfrac{750}{4\times 270}\\\Rightarrow L=0.69444\ m[/tex]

At f = 2300 Hz v = 750 m/s

[tex]L=\dfrac{750}{4\times 2300}\\\Rightarrow L=0.08152\ m[/tex]

At f = 270 Hz v = 350 m/s

[tex]L=\dfrac{350}{4\times 270}\\\Rightarrow L=0.32407\ m[/tex]

At f = 2300 Hz v = 350 m/s

[tex]L=\dfrac{350}{4\times 2300}\\\Rightarrow L=0.03804\ m[/tex]

Given temperature ϑ = 56°C. Formula: speed of sound c = 331.3 + 0.606 × ϑ.Atϑ = 56°C the speed of sound c = 331.3 + 0.606 × 56 = 365.2 m/s.

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

[tex]t_1 = \sqrt{\frac{2H}{g}}[/tex]

now the sound will come back to the observer in the time

[tex]t_2 = \frac{H}{v}[/tex]

so we will have

[tex]t_1 + t_2 = 2.42[/tex]

[tex]\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42[/tex]

so we have

[tex]\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42[/tex]

solve above equation for H

[tex]H = 26.8 m[/tex]

Part b)

If sound reflection part is ignored then in that case

[tex]H = \frac{1}{2}gt^2[/tex]

[tex]H = \frac{1}{2}(9.81)(2.42^2)[/tex]

[tex]H = 28.7 m[/tex]

so here percentage error in height calculation is given as

[tex]percentage = \frac{28.7 - 26.8}{26.8} \times 100[/tex]

[tex]percentage = 7.18[/tex]

365 meters/second

Explanation:

The speed of sound in air (v, in meters/second) is computed as function of temperature (T, in Celsius) by the following formula:

v = 331.4 + 0.6*T

Replacing with T = 56 °C we get:

v = 331.4 + 0.6*(56) = 365 meters/second

Ans : The speed of sound is the distance travelled per unit time by a sound wave as it propagates through an elastic medium. At 20 Â°C (68 Â°F), the speed of sound is 343 metres per second (1,125 ft/s; 1,235 km/h; 767 mph; 667 kn), or a kilometre in 2.91 s or a mile in 4.69 s. The speed of sound in water increases with increasing water temperature, increasing salinity and increasing pressure (depth). The approximate change in the speed of sound with a change in each property is: Temperature 1Â°C = 4.0 m/s Salinity 1PSU = 1.4 m/s Depth (pressure) 1km = 17 m/s

Given temperature ϑ = 56°C. Formula: speed of sound c = 331.3 + 0.606 × ϑ.Atϑ = 56°C the speed of sound c = 331.3 + 0.606 × 56 = 365.2 m/s.

In gases, the higher the speed of sound in that medium, the higher the pitch will be, when you sing. Only because of the decreasing air temperature, which decreases with altitude (height), the speed of sound decreases.