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Vinvika [58]
3 years ago
7

A yodeler yodels at a cliff 2000 m away. The sound of the echo returns in 9.8 s. If the wavelength of the sound wave is .85 m th

en what is the period of the sound wave?
Physics
1 answer:
stepladder [879]3 years ago
5 0

Answer:

Hey man, are you who I think you are? ... you used your student number on your account.... anyways im in your class :) guess who I am for free points. so the formula for period is Period=Wavelength/Velocity, so P=.85/velocity, to find velocity would be using the formula speed=distance/time, so the velocity is 2000/9.8 which is around 204.0816327.... so back to period formula, P=.85m/204.08 which gives a period of .004165. But im not entire sure so you have to check that on your own. (gl guessing who I am. hehe)

Explanation:

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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
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Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

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Then, the flow is said to be  in the opposite direction

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s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

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R = universal gas constant

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P_2 = pressure at location 2

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s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

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