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3 years ago

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Calculate the frequencies (in Hz) for the ten lowest modes of a rigid-wall room of dimensions 2.59m x 2.42m x 2.82m (i.e., find

the modes with the 10 lowest natural frequencies).
For each of these 10 natural frequencies, identify the value of nx, ny, and nz.

1 answer:

Digiron [165]3 years ago

3 0

Answer:

For This Answer Please See the Attached File.

Explanation:

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Crawling in heavy smoke or extreme heat can increase <u><em>visibility</em></u>

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4 years ago

What resources did Margaret Hutchinson Rousseau use ?

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Explanation:

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3 years ago

Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 65 and standard deviati

Sedbober [7]

Answer:

n this question, we are asked to find the probability that

R1 is normally distributed with mean 65 and standard deviation 10

R2 is normally distributed with mean 75 and standard deviation 5

Both resistor are connected in series.

We need to find P(R2>R1)

the we can re write as,

P(R2>R1) = P(R2-R1>R1-R1)

P(R2>R1) = P(R2-R1>0)

P(R2>R1) = P(R>0)

Where;

R = R2 - R1

Since both and are independent random variable and normally distributed, we can do the linear combinations of mean and standard deviations.

u = u2-u1

u = 75 - 65 = 10ohm

sd = √sd1² + sd2²

sd = √10²+5²

sd = √100+25 = 11.18ohm

Now we will calculate the z-score, to find P( R>0 )

Z = ( X -u)/sd

the z score of 0 is

z = 0 - 10/11.18

z= - 0.89

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4 years ago

Tasya [4]

Answer:

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2 years ago

Read 2 more answers

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

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4 years ago