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docker41 [41]
4 years ago
14

A chain saw produces a spherical sound wave having a frequency of 214Hz in air at 358C (308.2K or 958F). At a distance of 600mm

(23.62 in), the sound pressure level is 100 dB. Determine (a) the acoustic power level and (b) the rms acoustic particle velocity and velocity level at a distance of 600mm from the saw.
Physics
1 answer:
Helen [10]4 years ago
8 0

Given Information:

Frequency = 214 Hz

Temperature = 358° C = 308.2 K

Sound Pressure = p = 100 dB = 2 pascal

Distance = 600 mm = 0.60 m

Required Information:

(a) Acoustic Power Level = ?

(b) Acoustic Particle Velocity = ?

and Velocity level at a distance of 600 mm = ?

Answer:

Acoustic Power Level = Lw = 106.44 dB

Acoustic Particle Velocity = v = 0.0106 m/s

Velocity level = 60.25 dB

Explanation:

(a) Acoustic Power Level

Acoustic Power = W = 4πr² I

Acoustic Intensity = I =  p ²/Z₀

Where Z₀ is the characteristic impedance of air Z₀ = 409.8 rayl

I =  p ²/Z₀ = (2)²/409.8 = 0.00976 W/m²

W = 4πr² I = 4*π(0.60)²*0.00976 = 0.0441 W

Acoustic Power Level = Lw = 10log(W/Wref)

Where Wref is Reference Acoustic Power Wref = 1x10⁻¹² W

Lw = 10log(W/Wref) = 10log(0.0441/1x10⁻¹²) = 106.44 dB

Lw = 106.44 dB

(b) Acoustic Particle Velocity

Acoustic Particle velocity = v =  p ²/Zs

Where Zs is specific acoustic impedance

Zs =  Z₀kr/(1 + k²r²)⁰°⁵

Where k = 2πf/c and c = 346.1 m/s is the speed of sound in air

k = 2π*214/346.1 = 3.885 per m

Zs =  409.8*3.885*0.60/(1 + (3.885)²(0.60)²)⁰°⁵

Zs = 376.6 rayl

v =  p ²/Zs = 2²/376.6 = 0.0106 m/s

v = 0.0106 m/s

Velocity level = 10log(v/vref) where vref = 10x10⁻⁹ m/s

Velocity level = 10log(0.0106/10x10⁻⁹) = 60.25 dB

Velocity level = 60.25 dB

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Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

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#LearnwithBrainly

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A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
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This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

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where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

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a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

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(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

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Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

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We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

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(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

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The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

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v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

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m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

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mg = ma

which means that the acceleration is

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and the negative sign means it points downward.

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