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Softa [21]
2 years ago
9

Explain Sound level intensity with mathematical steps?

Physics
1 answer:
yan [13]2 years ago
8 0

Answer:

Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be β(dB)=10log10(II0)β(dB)=10log10⁡(II0), where I0 = 10−12 W/m2 is a reference intensity. In particular, I0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because β is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard (10−12 W/m2, in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.

Table 1. Sound Intensity Levels and IntensitiesSound intensity level β (dB)Intensity I(W/m2)Example/effect01 × 10–12Threshold of hearing at 1000 Hz101 × 10–11Rustle of leaves201 × 10–10Whisper at 1 m distance301 × 10–9Quiet home401 × 10–8Average home501 × 10–7Average office, soft music601 × 10–6Normal conversation701 × 10–5Noisy office, busy traffic801 × 10–4Loud radio, classroom lecture901 × 10–3Inside a heavy truck; damage from prolonged exposure[1]1001 × 10–2Noisy factory, siren at 30 m; damage from 8 h per day exposure1101 × 10–1Damage from 30 min per day exposure1201Loud rock concert, pneumatic chipper at 2 m; threshold of pain1401 × 102Jet airplane at 30 m; severe pain, damage in seconds1601 × 104Bursting of eardrums

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As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pres
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Answer:

Vi = 0.055 m³ = 55 L

Explanation:

From first Law of Thermodynamics, we know that:

ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

Vi = 0.058 m³ - 0.003 m³

<u>Vi = 0.055 m³ = 55 L</u>

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