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crimeas [40]
3 years ago
14

What is the speed of sound in dry air at 0°C? A. 316 meters/second B. 331 meters/second C. 346 meters/second D. 373 meters/secon

d
Physics
2 answers:
Katena32 [7]3 years ago
8 0

Answer:

1,450 meters per second

Explanation:

amm18123 years ago
3 0

Answer:

The speed of sound in dry air at 0°C is 331 meters/second

Explanation:

Sound wave is a longitudinal wave. The speed of sound depends on the density and compressibility of the medium. If the density of the medium is less, then it will have faster speed of sound and if the compressibility of the medium is high, it will have slower speed of sound.

It also depends on the temperature. At 20 degrees Celsius, the speed of sound is 343 m/s. At 25 degrees Celsius, its speed is 346 m/s and at 0 degrees Celsius, the speed of sound in dry air is 331.3 m/s

The formula is as follows :

v=(331.3+0.606\times T)\ m/s

T = 0°C

v = 331.3 m/s

or

v = 331 m/s

So, the correct option is (b) "331 meters/second".

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A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

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3 years ago
Relay of thermostate is not working why
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7 0
3 years ago
The relationship between distance from the sun and orbital period is that as the distance from the sun increases, the orbital pe
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3 years ago
An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd
Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ)  β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

    = 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

    = 1.07 × 10³(1 - [0.024]²)

    =  1.07 × 10³(1 - 0.000576)

    = 1.07 × 10³(0.999424)

    = 1069.38 gallons

7 0
4 years ago
How many electrons must be remowel from an electricaly Nurutral Silvor Coin to five it a charge of 3.2 NC ?
kakasveta [241]

Answer:

 #_electrons = 2 10¹⁰ electrons

Explanation:

For this exercise we can use a direct rule of three proportions rule. If an electron has a charge of 1.6 10⁻¹⁹ C how many electrons have a charge of 3.2 10⁻⁹ C

          #_electrons = 3.2 10⁻⁹ ( \frac{1}{1.6 \ 10^{-19}})

          #_electrons = 2 10¹⁰ electrons

7 0
3 years ago
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