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emmainna
11 days ago
7

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

servoir of the nozzle, the pressure and temperature are 10 atm and 300 K, respectively. At the nozzle exit, the pressure is 1 atm. Calculate the temperature and density of the flow at the exit. Assume that the flow is isentropic and (of course) compressible.
Physics
1 answer:
Softa [913]11 days ago
4 0

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Reservoir pressure = 10 atm

T_1 = Reservoir temperature = 300 K

P_2 = Exit pressure = 1 atm

T_2 = Exit temperature

R_s = Specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

Assuming isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

Flow density at exit measures 2.2721 kg/m³

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Yuliya22 [1153]

To tackle this question, we know the following:

1 Albert equals 88 meters.

1 A = 88 m.

Initially, we square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Thus, one acre is approximately 0.52 square alberts.</span>

7 0
3 days ago
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t
Softa [913]
The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "Doppler's effect."

The standard formula for the Doppler effect is:
f = (\frac{g + v_{r}}{g + v_{s}})f_{o} -- (A)

Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.

Where,
g = Speed of sound = 340m/s.
v_{r} = Velocity of the observer relative to the medium =?.
v_{s} = Velocity of the source in relation to the medium = 0 m/s.
f_{o} =  Frequency emitted from the source = 400 Hz.
f = Frequency recognized by the observer = 408 Hz.

Substituting the given values into equation (A) will yield:

408 = ( \frac{340 + v_{r}}{340 + 0})*400

\frac{408}{400} = \frac{340 + v_{r}}{340}

Solving the above will result in,
v_{r} = 6.8 m/s

The correct result = 6.8m/s



7 0
5 days ago
By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a
Ostrovityanka [942]

Broad questions addressed by conducting this experiment involve the effects of electric current.

Additional details

Electric current measures the quantity of electric charge passing per unit time.

It results from electrons moving due to a voltage difference (high potential to low potential) between two points.

These electrons flow through wires acting as conductors.

Ohm's Law states that:

The potential difference across a conductor is proportional to the current flowing through it, assuming resistance remains the same.

\displaystyle I=\frac{V}{R}\\\\V=I\times R

A basic electrical circuit consists of a voltage source (battery) and a lamp.

Ammeters used to measure current must be connected in series with the load.

By adjusting the voltage while resistance is constant, varying current values are observed; increasing voltage produces higher current.

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Electron flow inside devices

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Keywords: basic electric circuits, Ohm's law, experiment

5 0
13 days ago
Read 2 more answers
After soccer practice coach Miller goes to the roof of the school to retrieve the event soccer balls the height of the school is
Yuliya22 [1153]

Answer:

50.2 cm

Explanation:

We have the following data:

Height, h=3.5 m

Initial horizontal velocity, u_x=15 m/s

Time, t=0.32 s

We need to determine how far the ball is from the ground after 0.32 s.

Initial vertical velocity, u_y=0

s=u_yt+\frac{1}{2}gt^2

Where g=9.8 m/s^

s=0+\frac{1}{2}(9.8)(0.32)^2

s=0.502 m

s=0.502\times 100=50.2 cm

4 0
6 days ago
A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine
Sav [1105]

Given


m1(mass of red bumper): 225 Kg


m2 (mass of blue bumper): 180 Kg


m3(mass of green bumper): 150 Kg


v1 (velocity of red bumper): 3.0 m/s


v2 (final velocity of the combined bumpers):?




The principle of momentum conservation indicates that the momentum before impacts equals the momentum after impacts. This can be represented mathematically as:


Pa= Pb


Pa symbolizes the momentum prior to collision and Pb refers to momentum after collision.


Applying this principle to the aforementioned scenario results in:


Momentum pre-collision= momentum post-collision.


Momentum pre-collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s


Momentum post-collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

We now know that Momentum pre-collision equals momentum post-collision.


<presulting in="">

1215 = 555 v2


v2 = 2.188 m/s


Consequently, the final velocity of the combined bumper cars is 2.188 m/s

</presulting>
4 0
3 days ago
Read 2 more answers
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