Please help meh!!! I appreciate it!

A flame ionization detector, which is often used in gas chromatography, responds to a change in

Lena [83]

Answer:

Option A

Explanation:

We know that ions are present in hydrogen-air flame and when the burning of an organic compound takes place in this flame more ions are produced in the flame.

Thus when we apply a voltage across this flame, the ion collector plate attracts the all the ions in the flame.

The presence of organic compounds increases the voltage across the hydrogen ion flame produced at the ion collector increases and as the voltage increases, the detection of the organic compound can be made in turn.

Thus flame ionization detector clearly responds to the variation in the collection of ions or electrons in a flame.

3 0

3 years ago

Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is

GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

public static void main(String[] args) {

final int MAXCHEESE = 10;

String[] names = new String[MAXCHEESE];

double[] prices = new double[MAXCHEESE];

double[] amounts = new double[MAXCHEESE];

// Three Special Cheeses

names[0] = "Humboldt Fog";

prices[0] = 25.00;

names[1] = "Red Hawk";

prices[1] = 40.50;

names[2] = "Teleme";

prices[2] = 17.25;

System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

System.out.println(names[0] + ": $" + prices[0] + " per pound");

System.out.println(names[1] + ": $" + prices[1] + " per pound");

System.out.println(names[2] + ": $" + prices[2] + " per pound");

Random ranGen = new Random(100);

// error at initialising i

// i should be from 0 to MAXCHEESE value

for (int i = 0; i < MAXCHEESE; i++) {

names[i] = "Cheese Type " + (char) ('A' + i);

prices[i] = ranGen.nextInt(1000) / 100.0;

amounts[i] = 0;

System.out.println(names[i] + ": $" + prices[i] + " per pound");

}

7 0

3 years ago

You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-

Katen [24]

Answer:

The required wall thickness is $1.506 \times 10^{-3}$ m

Explanation:

Given:

Fluid density $\rho = 1200$ $\frac{kg}{m^{3} }$

Diameter of tank $d = 8$ m

Length of tank $l = 4$ m

F.S = 4

For A-36 steel yield stress $\sigma = 250$ MPa,

Allowable stress $\sigma _{allow} = \frac{\sigma}{F.S}$

$\sigma _{allow} = \frac{250}{4} = 62.5$ MPa

Pressure force is given by,

$P = \rho gh$

$P = 1200 \times 9.8 \times 4$

$P = 47088$ Pa

Now for a vertical pipe,

$\sigma _{allow} = \frac{Pd}{4t}$

Where $t =$ required thickness

$t = \frac{Pd}{4 \sigma _{allow} }$

$t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }$

$t = 1.506 \times 10^{-3}$ m

Therefore, the required wall thickness is $1.506 \times 10^{-3}$ m

8 0

3 years ago

A wastewater treatment plant has two primary clarifiers, each 20m in diameter with a 2-m side-water depth. the effluent weirs ar

jasenka [17]

Answer:

overflow rate 20.53 m^3/d/m^2

Detention time 2.34 hr

weir loading 114.06 m^3/d/m

Explanation:

calculation for single clarifier

$sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d$

$surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2$

$surface area = 314.16 m^2$

volume of tank $V = A\times side\ water\ depth$

$=314.16\times 2 = 628.32m^3$

$Length\ of\ weir = \pi \times diameter of weir$

$= \pi \times 18 = 56.549 m$

overflow rate = $v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2$

Detention time $t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr$

weir loading $= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m$

6 0

3 years ago

3.

Andreyy89

Answer:

7

Explanation:

5 + 2 = 7

4 0

2 years ago