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riadik2000 [5.3K]

2 years ago

15

Your friend has two substances A and B which are compressed liquid and superheated vapor respectively. Both are in rigid vessels

. Your friend claims that when he cools down both substances they started to boil. Can this be true? Explain yourself using a T-v diagram.

1 answer:

Anon25 [30]2 years ago

3 0

Answer:

Following are the solution to the given question:

Explanation:

The material is necessary to cook since frying is a speedy process for evaporation.

Drug A is now in the compressed fluid area, and the material would not boil if the pressure is chilled. Because the ship is solid, its substance A claim is false.

Unless the volume comprises of a drop in the heat, the B substance reaches a vapor pressure area and a wet region. That's the area that melting may occur. His claim for material B could therefore be true.

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A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul

lorasvet [3.4K]

<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

$\frac{1}{R_x}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$

Solving for Rₓ gives;

$R_{x}$ = $\frac{R_1 * R_2}{R_1 + R_2}$ ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ = resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;

$R_{x}$ = $\frac{40 * 60}{40 + 60}$

$R_{x}$ = $\frac{2400}{100}$

$R_{x}$ = 24 Ω

Therefore, the total resistance is 24Ω

4 0

2 years ago

In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini

storchak [24]

Answer:

attached below

Explanation:

8 0

2 years ago

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

vekshin1

Solution :

<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>

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 <u> 6.3 </u> 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

The yield of a chemical process is being studied.The two most important variables are thought to be the pressure and the tempera

anyanavicka [17]

Answer:

<u>note:</u>

<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</u>

7 0

2 years ago

Solve using Matlab the problems:

Firlakuza [10]

Answer:

Explanation:

% Clears variables and screen

clear; clc

% Asks user for input

n = input('Total number of objects: ');

r = input('Size of subgroup: ');

% Computes and displays permutation according to basic formulas

p = 1;

for i = n - r + 1 : n

p = p*i;

end

str1 = [num2str(p) ' permutations'];

disp(str1)

% Computes and displays combinations according to basic formulas

str2 = [num2str(p/factorial(r)) ' combinations'];

disp(str2)

=================================================================================

Example: check

How many permutations and combinations can be made of the 15 alphabets, taking four at a time?

The answer is:

32760 permutations

1365 combinations

==================================================================================

7 0

2 years ago