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Natali5045456 [20]
3 years ago
10

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any wo

rk or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30°C enters it with a velocity of 356 m/s and the exit state is 200 kPa and 90°C. The specific heat of air at the average temperature of 60°C
Physics
1 answer:
Mice21 [21]3 years ago
6 0

Answer:

The exit velocity air is 77.56 m/s.

Explanation:

Specific heat of air at 60°C is about 1.006 KJ/kg-K (from standard table).

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

For ideal gas h=CT

Given that T_1=30°C

T_2=90°C

V_1=356 m/s

By putting the values

1.006\times (273+30)+\dfrac{356^2}{2000}=1.006\times (273+90)+\dfrac{V_2^2}{2000}+

V_2=77.56 m/s

So the exit velocity air is 77.56 m/s.

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Which of these would best describe the outcome of increasing the kinetic energy of the molecules in the above picture? A) The te
VikaD [51]

Answer: Option (A) is the correct answer.

Explanation:

Relation between kinetic energy and temperature is as follows.

                K.E \propto \frac{3}{2}kT

As, kinetic energy is directly proportional to temperature. So, when there will be increase in temperature then there will also occur increase in kinetic energy of the particles of a substance.

And, when gas particles move slowly then it means kinetic energy of gas particles is very low. It also implies that temperature is low.

Thus, we can conclude that temperature will increase best describes the outcome of increasing the kinetic energy of the molecules in the above picture.

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3 years ago
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Write the factors of potential energy
IgorLugansk [536]
It depends on two factors height and mass

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3 years ago
An element has three stable isotopes. One has a mass number of 63 and an abundance of 50%. The second has a mass number of 65 an
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63x1/2 + 65x0.3 + 67x1/5

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4 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

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3 years ago
A dipole with a positive charge of 2.0 uC and a negative charge of -2 uC is centered at the origin and oriented along the x axis
LuckyWell [14K]

Answer:

The reulting electric field at x = 4.0 and y = 0.0 from the dipole is 0.5612 N/C

Solution:

As per the question:

Charges of the dipole, q = \pm 2\mu C

Separation distance between the charges, d = 0.0010 m

Separation distance between the center and the charge, d' = \frac{d}{2} = 5\times 10^{- 4} m

x = 4.0 m

y = 0.0 m

Now,

The electric field due to the positive charge on the right of the origin:

E = k\frac{q}{(d' + x)^{2}}

where

k = Coulomb's constant = 9\times 10^{9} Nm^{2}C^{- 2}

Now,

E = (9\times 10^{9})\frac{2\times 10^{- 6}}{(5\times 10^{- 4} + 4)^{2}} = 1124.72\ N/C

Similarly, electric field due to the negative charge:

E' = k\frac{q}{(x - d')^{2}}

E' = (9\times 10^{9})\frac{2\times 10^{- 6}}{(4 - 5\times 10^{4})^{2}} = - 1125.28\ N/C

Thus

E_{total} = E' - E = 0.5612 N/C

8 0
4 years ago
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