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3 years ago

Can someone help with these questions (1 , 3 , 4)

Physics

1 answer:

LenaWriter [7]3 years ago

8 0

Remark

4 is long (but not hard) and I am in a terrible time bind. I will take a short at doing it but it will be abbreviated.

One

Ax = 70*cos(51.7) = 43.38

Ay = 70*sin(51.7) = 54.93

That makes C the correct answer. To resolve these, Ax is always taken along the +x axis. and Ay is always associated with the y axis. A(x) always uses Cos(x) and Ay = sin(x)

Three

A would be true if Ax and Ay were vectors. They don't seem to be. They look like scalers. Magnitudes don't add that way.

B Theta = tan-1(Ay/Ax) The Statement given is simply not true.

C Tan(Theta) = Ay / Ax Not The given result

D is correct Cos(theta) = The magnitude along the +x axis divided by the hypotenuse which is sqrt(Ax^2 + Ay^2)

Four

The best way to do this is to set up x and y values in a table. I can't do this well, but I'll try.

x y

12*cos(53) = 7.22 15*sin(53)= 9.58

15*cos(-40) = 11.49 15.sin(-40) = -9.64

Sum = 18.69 = - 0.08

As you can see most of the magnitude and direction will come from Ax

So the answer is 19 for the magnitude and 0o for the direction. The other choices are taken from theta = 40o. That minus sign makes all the difference.

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3 years ago

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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

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Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

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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

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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:

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

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

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

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

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