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prisoha [69]
3 years ago
6

Please help.

Physics
1 answer:
cestrela7 [59]3 years ago
3 0
Hello,

<span>A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and returns with a frequency of 1250 Hz. 

How fast is the vehicle going?

Doppler equation formula: </span>ƒL = ƒS(v - vL)/(v - vS)

The wave returns with a frequency of 1250 Hz, the <span>echo frequency is higher; the car must be traveling towards the police car. 

</span><span>The wave echo is coming back towards the police car at the same speed as the sound wave travels towards the moving car so t</span><span>he relative speed between the cars is half of the speed of the echo.

* </span><span>speed of sound equals about 337 m/s </span>

2v / 337 = (1250/1200) - 1 
<span>2v = 14.04 m/s </span>
<span>v = 7.02 m/s
</span>
Thus, the vehicle is going 7.02 m/s.

Faith xoxo

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The correct answer is B.

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Which of the following BEST describes the term "air resistance"?
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A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

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Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

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h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

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3 years ago
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