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Varvara68 [4.7K]

2 years ago

10

The three sub regions of South America are the Andes Mountains, the Amazon Rainforest, and the Eastern Highlands. The Atacama De

sert is the driest place on Earth.

2 answers:

Leto [7]2 years ago

8 0

Answer:

<:

Explanation:

pentagon [3]2 years ago

4 0

Answer:

d

Explanation:

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Cuál de las siguientes es la mejor manera de practicar sus habilidades de tecnología de secundaria?

Mkey [24]

Huh? Do you know English?

8 0

3 years ago

A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,

gulaghasi [49]

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

8 0

3 years ago

In an ideal gas, specific enthalpy is a function of i. Entropy ii. Temperature iii, Pressure iv. Mass

Mice21 [21]

Answer:

Temperature

Explanation:

In an ideal gas the specific enthalpy is exclusively a function of Temperature only this can be also written as h = h(T)

A gas is said be ideal gas if obeys PV= nRT law

And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.

5 0

3 years ago

A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop o

anastassius [24]

Answer:

1500Ω

Explanation:

Given data

voltage = 15 V

total Resistance = 4000Ω

potential drop V = 9.375 V

To find out

R2

Solution

we know R1 +R2 = 4000Ω

So we use here Ohm's law to find out current I

current = voltage / total resistance

I = 15 / 4000 = 3.75 × $10^{-3}$ A

Now we apply Kirchhoffs Voltage Law for find out R2

R2 = ( 15 - V ) / current

R2 = ( 15 - 9.375 ) / 3.75 × $10^{-3}$

R2 = 1500Ω

6 0

3 years ago

Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly

-Dominant- [34]

Answer:

The answer is " $3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}$ "

Explanation:

Given data:

Initial temperature of tank $T_1 = 300^{\circ}\ C= 573 K$

Initial pressure of tank $P_1= 400 \ kPa$

Diameter of throat $d* = 2 \ cm$

Mach number at exit $M = 2.8$

In point a:

calculating the throat area:

$A*=\frac{\pi}{4} \times d^2$

$=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2$

Since, the Mach number at throat is approximately half the Mach number at exit.

Calculate the Mach number at throat.

$M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4$

Calculate the exit area using isentropic flow equation.

$\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}$

Here: $\gamma$ is the specific heat ratio. Substitute the values in above equation.

$\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm$

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

$\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K$

Calculate the velocity at exit.

$V*=M*\sqrt{ \gamma R T*}$

Here: R is the gas constant.

$V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}$

Calculate the density of air at inlet

$\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}$

Calculate the density of air at throat using isentropic flow equation.

$\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}$

Calculate the mass flow rate.

$m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}$

5 0

2 years ago