All categories

2 years ago

What is FOLLOWING PROCEDURES

Engineering

1 answer:

il63 [147K]2 years ago

4 0

Answer:

When employees follow procedures, they perform tasks correctly and provide consistent customer service. This enhances the quality of your organization's products and services. And, in turn, improves your company's reputation. Employees can know they are fulfilling their roles and take pride in their work.

You might be interested in

Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam

Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

Rod 1 : P*cos(Q)

Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

Q = arctan ( y_1 / x_1)

Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

Rod 1 : stress_1 = P*cos(Q) / A_1

Rod 2: stress_2 = - P*sin(Q) / A_2

- The cross sectional Area of both rods are A_1 and A_2:

A_1 = pi*D^2 / 4

A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

Rod 1: P_max = stress_1 * A_1 / cos(Q)

P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

P_max = 25203.61848 N

Rod 2: P_max = stress_2 * A_2 / sin(Q)

P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

8 0

2 years ago

A storm sewer is carrying snow melt containing 1.2 g/L of sodium chloride into a small stream. The stream has a naturally occurr

galina1969 [7]

Answer:

Given Data:

concentration of sewer Csewer = 1.2 g/L

converting into mg/L = Csewer = 1.2 g/L x 1000 mg/g = 1200 mg/L

flow rate of sewer Qsewer = 2000 L/min

concentration of sewer Cstream = 20 mg/L

flow rate of sewer Qstream = 2m3/s

converting Q into L/min = 2m3/s x 1000 x 60 = 120000 L/min

mass diagram is

6 0

2 years ago

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb

Minchanka [31]

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

= √( 140*2 ) / ( 12 * 10^-3 * 55 )

= √ 280 / 0.66 = 20.60

Attached below is the remaining answers

7 0

2 years ago

Reduce the force F ij = + (2 5 ) kN to point A(2m,3m) that acts on point B( 3m,5m) - .

Alexxx [7]

Given :

Force, $\vec{F}= (2\hat{i} + 5\hat{j})\ kN$ .

Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )

To Find :

The work done by force F .

Solution :

Displacement vector between point A and B is :

$\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}$

Now, we know work done is given by :

$W = \vec{F}.\vec{d}\\\\W= (2\hat{i} + 5\hat{j}).(\hat{i}+\hat{2j})\\\\W = (2\times 1) +( 5\times 2) \ kJ\\\\W = 12 \ kJ$

W = 12000 J

Therefore, work done by force is 12000 J .

6 0

2 years ago

Write a program that prompts the user to enter time in 12-hour notation. The program then outputs the time in 24-hour notation.

Juliette [100K]

Answer:

THE CODE FOR THE PROGRAM IS GIVEN BELOW:

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int main()

{

convertTime convert;

int hr, mn, sc = 0;

cout << "Please input hours in 12 hr notation: ";

cin >> hr;

cout << "Please input minutes: ";

cin >> mn;

cout << "Please input seconds: ";

cin >> sc;

convert.invalidHr(hr);

convert.invalidMin(mn);

convert.invalidSec(sc);

convert.printMilTime();

system("Pause");

return 0;

}

#include <iostream>

#include "ConvertTimeHeader.h"

using namespace std;

int convertTime::invalidHr (int hour)

{

try{

if (hour < 13 && hour > 0)

{hour = hour + 12;

return hour;}

else{

cin.clear();

cin.ignore();

cout << "Invalid input! Please input hour again in correct 12 hour format: ";

cin >> hour;

invalidHr(hour);

throw 10;

}

catch (int c) { cout << "Invalid hour input!";}

}

int convertTime::invalidMin (int min)

{

try{

if (min < 60 && min > 0)

{return min;}

else{

cin.clear();

cin.ignore();

cout << "Invalid input! Please input minutes again in correct 12 hour format: ";

cin >> min;

invalidMin(min);

throw 20;

return 0;

}

catch (int e) { cout << "Invalid minute input!" << endl;}

}

int convertTime::invalidSec(int sec)

{

try{

if (sec < 60 && sec > 0)

{return sec;}

else{

cin.clear();

cin.ignore();

cout << "Invalid input! Please input seconds again in correct 12 hour format: ";

cin >> sec;

invalidSec(sec);

throw 30;

return 0;

}

catch (int t) { cout << "Invalid second input!" << endl;}

}

void convertTime::printMilTime()

{

cout << "Your time converted: " << hour << ":" << min << ":" << sec;

}

Explanation:

4 0

3 years ago