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Mama L [17]
1 year 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:
Lady_Fox [76]1 year ago
4 0

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For 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

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\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

The density of the flow at the exit is 2.2721 kg/m³

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4 0
1 year ago
An experiment is conducted in which red light is diffracted through a single slit. Listed below are alterations made, one at a t
Xelga [282]

Answer:

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

Explanation:

As we know that the position of first minimum is given as

a sin\theta = N\lambda

so we have

\theta = sin^{-1}(\fracN\lambda}{a})

so width of minimum is given as

w = L\times sin^{-1}(\fracN\lambda}{a})

now if we need to decrease the angular position of minimum

1). so we can decrease the distance of screen from the slit

2). we can decrease the wavelength

3). We can increase the width of the slit

So correct answer will be

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

6 0
2 years ago
Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.
VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0
1 year ago
A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

4 0
1 year ago
What principles of science (like facts, laws, and theories) might help explain why similar investigations conducted in many part
natulia [17]

Answer:

Reproducibility of research

Explanation:

The principle of science that explains why similar experimental investigations conducted in different parts of the world could result in the same outcome is referred to as reproducibility.

<em>A good research or experiment in science must be reproducible, otherwise, the outcome of such an experiment might become inadmissible within the scientific community. It is a core principle of the scientific method that similar results should be obtained when an experiment or observational study conducted in one place is repeated in another place with the same procedure. Hence, an experiment must be reproducible in science in order for the outcome of such an experiment to be part of the general scientific knowledge. </em>

7 0
1 year ago
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