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Ahat [919]
4 years ago
5

(20 points) (Assessment of Outcome 1): A plant has two identical standby generator units for emergency use. In the area of the g

enerators, the normal noise level registers 82 dBA on the sound-level meter with the generators turned off. When one generator switches on, the SLM needle jumps to 85.8 dBA. What will the dBA reading be when the second generator also turns on (so that both generators are on)
Engineering
1 answer:
erastovalidia [21]4 years ago
5 0

Answer:

It wouldn't get any louder then maybe 3db more

Explanation:

There's even a equation if you wanted to check this out but, if they are the same generator same model and all and made the same precise noise it wouldn't increase more then 3db.

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Answer:By obey peoples propertice

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8 0
2 years ago
Nitrogen at an initial state of 300 K, 150 kPa and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 8
s344n2d4d5 [400]

Answer:

Work = - 4175.23 J

Heat Transfer = -4175.23 J

Explanation:

The work done in an isothermal process, is given by the following formula:

W = RT ln (P1/P2)

where,

W = Work done

R = Universal Gas Constant = 8.314 J/mol.k

T = Constant Temperature = 300 K

P1 = initial pressure = 150 KPa

P2 = final pressure = 800 KPa

using these values, we get:

W = (8.314 J/mol.K)(300 k) ln (150/800)

<u>W = - 4175.23 J</u>

Here, negative sign shows that work is done on the system.

In isothermal process, from 1st law of thermodynamics:

Heat Transfer = Q = W     (Since change in internal energy is zero for isothermal processes)

<u>W = - 4175.23 J</u>

Here, negative sign shows that heat is transferred from the system to surrounding.

6 0
3 years ago
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Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e
marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

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We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

\Delta x=\frac{PL^3}{3EI}

Hence we can write

P=\frac{3EI\cdot \Delta x}{L^3}

Comparing with the standard spring equation F=kx we find the cantilever analogous to spring with k=\frac{3EI}{L^3}

Now the angular frequency of a spring is given by

\omega =\sqrt{\frac{k}{m}}

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Thus applying values we get

\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}

\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec

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3 years ago
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Answer:

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Explanation:

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Answer:

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