A set of bus bars also can be called a

Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.

vekshin1

Solution :

Sieve Size (in) Weight retain(g)

3 1.62

2 2.17

$1\frac{1}{2}$ 3.62

$\frac{3}{4}$ 2.27

$\frac{3}{8}$ 1.38

PAN 0.21

Given :

Sieve weight % wt. retain % cumulative % finer

size retained wt. retain

No. 4 59.5 10.225% 10.225% 89.775%

No. 8 86.5 14.865% 25.090% 74.91%

No. 16 138 23.7154% 48.8054% 51.2%

No. 30 127.8 21.91% 70.7154% 29.2850%

No. 50 97 16.6695% 87.3849% 12.62%

No. 100 66.8 11.4796% 98.92% 1.08%

Pan 6.3 1.08% 100% 0%

581.9 gram

Effective size = percentage finer 10% ( $$$D_{20}$ )

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x , 10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm) , 51.2%

N 8 (2.38 mm) , 74.91%

x, 60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

$Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4 and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

7 0

2 years ago

A 50 (ohm) lossless transmission line is terminated in a load with impedance Z= (30-j50) ohm. the wavelength is 8cm. Determine:

Serjik [45]

Answer:

Reflection Coefficient = $0.57e^{-i79.8}$

SWR=3.65

Position of $V_{max} =3.11cm$

position of $i_{max} =1.11cm$

Explanation:

To determine the above answers, let outline the useful formulas

refection coefficient $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$ .

where terminal impednce = (30-i50)Ω

characteristics impedance= 50Ω

Secondly, the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

Now let us substitute values and solve,

a. $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$

$p=\frac{(30-i50)-50}{(30-i50)-50} \\$

$p=\frac{-20-i50}{80-i50} \\$

multiplying the numerator and denominator by the conjugate of the denominator. we have

$p=\frac{-20-i50}{80-i50}*\frac{80+i50}{80+i50}\\$

by carrying out careful operation, we arrived at

$p=\frac{900-i5000}{8900} \\p=0.1011-i0.56179\\$

To express in polar form i.e $re^{i alpha}$

$r=\sqrt{0.1011^{2}+0.56179^{2}} \\r=0.57\\$

to get the angle

alpha= $tan^{-1} \frac{0.56179}{0.1011} \\alpha=-79.8\\$

hence the Reflection Coefficient,p = $0.57e^{-i79.8}$

b. we now determine the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

$swr=\frac{1+0.57}{1-0.57} =3.65\\$

c. to determine the position of the maximum voltage nearest to the load,

we use the equation

$Position of V_{max}=\frac{\alpha λ}{4\pi}+\frac{λ}{2}\\$

were $λ$ is the wavelength of 8cm

lets convert α to rad by multiplying by π/180

$Position of V_{max}=\frac{-79.8 *8cm*\pi}{4\pi*180 } +\frac{8cm}{2} \\$

$Position of V_{max}=-0.89+4.0=3.11cm\\$ .

d. also were we have minimum voltage,there the maximum current will exist, to find this position nearest to the load

$Position of v_{min}=Position of V_{max}-\frac{λ}{4} \\\\Position of v_{min}=3.11cm-\frac{8cm}{4}=1.11cm$ .

since the voltage minimum occure at 1.11cm. we can conclude that the current maximum also occur at this point i.e 1.11cm

3 0

3 years ago

Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00

fgiga [73]

Answer:

Here is the code.

Explanation:

#include<stdio.h>

int main()

{

int i, time_worked, over_time, overtime_pay = 0;

for (i = 1; i <= 10; i++)

{

printf("\nEnter the time employee worked in hr ");

scanf("%d", &time_worked);

if (time_worked>40)

{

over_time = time_worked - 40;

overtime_pay = overtime_pay + (12 * over_time);

}

printf("\nTotal Overtime Pay Of 10 Employees Is %d", overtime_pay);

return 0;

}

Output :

Enter the time employee worked in hr 42

Enter the time employee worked in hr 45

Enter the time employee worked in hr 42

Enter the time employee worked in hr 41

Enter the time employee worked in hr 50

Enter the time employee worked in hr 51

Enter the time employee worked in hr 52

Enter the time employee worked in hr 53

Enter the time employee worked in hr 54

Enter the time employee worked in hr 55

Total Overtime Pay Of 10 Employees Is 1020.

6 0

2 years ago

A contractor is planning on including several skylights in each unit of a residential development. What type of worker would she

Sladkaya [172]

Answer:

Glazier

Explanation:

Glaziers are workers who specializes in cutting and installation of glass works.

They work with glass in various surfaces and settings, such as cutting and installing windows and doors, skylights, storefronts, display cases, mirrors, facades, interior walls, etc.

Thus, the type of worker the contractor will hire for this project is a Glazier

8 0

3 years ago

Drag each tile to the correct box.

Trava [24]

Answer:

Bluray

DVD

CD

Explanation:

Blu ray can hold 25gb per layer

Dvd can hold 4.7GB on a single layer

Cd can hold around 737 mb

Also, dvds can go up to 2 layers

Blu ray can go up to 4

6 0

3 years ago