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Ad libitum [116K]
3 years ago
8

Which of the following describes how a sound wave can bounce off a surface?

Physics
2 answers:
Rainbow [258]3 years ago
5 0

Answer: reflection

Explanation:

Gennadij [26K]3 years ago
4 0

Answer:

a

Explanation:

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(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,
krek1111 [17]

Answer:

V_2 = 45.44m/s

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K

When we have all this values in SI apply a Energy Balance Equation,

\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0

Solving for V_2

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

From the table of gas properties we calculate for T_1 = 216K and T_2 = 250K

h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})

h_1 = 215.97kJ/kg

For T_2;

h_2 = 250.05kJ/kg

Substituting in equation for V_2

V_2 = \sqrt{V_1^2+2(h_1-h_2)}

V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s

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3 years ago
You estimate the radius of the big wheel to be 19 m
posledela
Explanation:

first, find the circumference of the wheel by using the formula 2(pi)(r):

2(pi)(19) = 119.380521

divide by 25 secs

119.380521/25 = 4.77522083

round to the nearest tenth is 4.8, so the speed is 4.8mm/sec
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