I feel so pressured..

generally compound curves are not filtered recommended for A. Road B. water way C. underground road D. rail way

vfiekz [6]

Answer:

C. underground road

Explanation:

Generally compound curves are not filtered and recommended for use in an underground road. However, they are best used in the road, water way, and rail way.

3 0

3 years ago

What is the locating position of the land field?

Ivahew [28]

Any point on earth can be located by specifying its latitude and longitude, including Washington, DC, which is pictured here. Lines of latitude and longitude form an imaginary global grid system, shown in Fig. 1.17. Any point on the globe can be located exactly by specifying its latitude and longitude.

4 0

2 years ago

Where has the process of nuclear fusion been occurring for over four billion years

dezoksy [38]

Answer:

Inside the Sun.

Explanation:

Inside the Sun, this process begins with protons (which is simply a lone hydrogen nucleus) and through a series of steps, these protons fuse together and are turned into helium. This fusion process occurs inside the core of the Sun, and the transformation results in a release of energy that keeps the sun hot.

8 0

3 years ago

The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8

il63 [147K]

Answer:

a.)

US Sieve no. % finer (C₅ )

4 100

10 95.61

20 82.98

40 61.50

60 42.08

100 20.19

200 6.3

Pan 0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no. Mass of soil retained (C₂ )

4 0

10 18.5

20 53.2

40 90.5

60 81.8

100 92.2

200 58.5

Pan 26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂× $\frac{100}{w}$

∴ we get

US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)

4 0 0

10 4.39 4.39

20 12.63 17.02

40 21.48 38.50

60 19.42 57.92

100 21.89 79.81

200 13.89 93.70

Pan 6.30 100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no. Cummulative % retained (C₄) % finer (C₅ )

4 0 100

10 4.39 95.61

20 17.02 82.98

40 38.50 61.50

60 57.92 42.08

100 79.81 20.19

200 93.70 6.3

Pan 100 0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = $\frac{D60}{D10}$

⇒ Cu = $\frac{0.4}{0.12} = 3.33$

d.)

Coefficient of Graduation = Cc = $\frac{D30^{2}}{D10 . D60}$

⇒ Cc = $\frac{0.22^{2}}{(0.4) . (0.12)}$ = 1

3 0

2 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