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scZoUnD [109]
3 years ago
5

A woman is sitting at a bus stop when an ambulance with a siren wailing at 317 Hz approaches at 69 miles per hour (mph). Assume

the speed of sound to be 343 m/s. a) How fast is the ambulance moving in meters per second? (perform the necessary unit conversion) Vs= 69 mph = m/s b) What frequency does the woman hear? fa = Hz c) What speed (vs) would the ambulance be traveling in order for the woman to hear the siren at an approaching frequency of 350 Hz? Vs= m/s d) What frequency would she hear as the siren moves away from her at the same speed (as in part c)? fa = Hz
Physics
1 answer:
xxMikexx [17]3 years ago
3 0

Answer:

a) 30.84m/s

b) 348.32Hz

c) 32.34m/s

d) 289.69Hz

Explanation:

a) If 1 mile=1609,34m, and 1 hour=3600 seconds, then 69mph=69*1609.34m/3600s=30.84m/s

b) Based on Doppler effect:

/*I will take as positive direction the vector \vec r_{observer}-\vec r_{emiter} */

f_{observed}=(\frac{v_{sound}-v_{observed}}{v_{sound}-v_{emited}})f_{emited}

f_{observed}=(\frac{343m/s-0m/s}{343m/s-30.84m/s})317Hz=348.32Hz

c) 350Hz=(\frac{343m/s-0m/s}{343m/s-v_{ambulance}})317Hz, V_{ambulance}=343m/s-\frac{317Hz}{350Hz}.343m/s=32.34m/s

d) f_{observed}=(\frac{343m/s-0m/s}{343m/s+32.34m/s})317Hz=289.69Hz

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3 years ago
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

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I hope it helps you!                                                                  

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Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

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1)

Lembre-se:

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2)

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