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3 years ago

5

Based on the concept that it is better to prevent falls happening in the first place, which of the following safety methods meet

s that criteria?

1 answer:

alekssr [168]3 years ago

7 0

Answer:fall arrest harness

Explanation:cuz it’s just right

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In part A you are asked to write the pseudocode for the program. In part B you are asked to write the syntax of the code for the

Naya [18.7K]

Answer:

C++.

Explanation:

#include <iostream>

#include <string>

using namespace std;

///////////////////////////////////////////////////////////////

int main() {

string quote, book;

int page;

cout<<"What is your favorite quote from a book?"<<endl;

getline(cin, quote);

cout<<endl;

/////////////////////////////////////////////

cout<<"What book was that quote from?"<<endl;

getline(cin, book);

cout<<endl;

/////////////////////////////////////////////

cout<<"What page was that quote from?"<<endl;

cin>>page;

cout<<endl;

/////////////////////////////////////////////

int no_of_upper_characters = 0;

for (int i=0; i<quote.length(); i++) {

if (isupper(quote[i]))

no_of_upper_characters++;

}

cout<<"No. of upper case characters: "<<no_of_upper_characters<<endl;

/////////////////////////////////////////////

int no_of_characters = quote.length();

cout<<"No. of characters: "<<no_of_characters<<endl;

/////////////////////////////////////////////

bool isDog = false;

for (int i=0; i<quote.length(); i++) {

if (isDog == true)

break;

else if (quote[i] == 'd') {

for (int j=i+1; j<quote.length(); j++) {

if (isDog == true)

break;

else if (quote[j] == 'o') {

for (int z=j+1; z<quote.length(); z++) {

if (quote[z] == 'g') {

isDog = true;

break;

}

}

}

}

}

}

if (isDog == true)

cout<<"This includes 'd' 'o' 'g' in the quote";

//////////////////////////////////////////////

return 0;

}

3 0

3 years ago

Who works alongside and assists the engineers?

nika2105 [10]

Answer:

<u>Assistants</u><u> </u><u>works alongside and assists the engineers.</u>

5 0

3 years ago

Technician A says high resistance causes an increase in current flow. Technician B says a higher than

Kryger [21]

The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.

<h3>Ohm's law</h3>

Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as

V = IR

Where

V is the potential difference

I is the current

R is the resistance

<h3>Technician A</h3>

High resistance causes an increase in current flow

V = IR

Divide both side by I

R = V / I

Thus, technician A is wrong as high resistance suggest low current flow

<h3>Technician B</h3>

Higher than normal voltage drop could indicate high resistance

V = IR

Thus, technician B is correct as high voltage indicates high resistance

<h3>Conclusion </h3>

From the above illustration, we can see that technician B is correct

Learn more about Ohm's law:

brainly.com/question/796939

8 0

2 years ago

For a project in C++ we are supposed toDesign a class named Month. The class should have the following private members:-name: a

mart [117]

Answer:

include <iostream>

using namespace std;

class Month

{

public:

Month (char firstLetter, char secondLetter, char thirdLetter);

Month (int monthNum);

.

Month();

void outputMonth_num();

void outputMonthLetters();

private:

int month;

};

int main ()

{

//

// Variable declarations

//

int monthNum;

char firstLetter, secondLetter, thirdLetter;

char testAgain;

do {

cout << endl;

cout << "Testing the default constructor ..." << endl;

Month defaultMonth;

defaultMonth.outputMonth_num();

defaultMonth.outputMonthLetters();

//

// Construct a month using the constructor with one integer argument

//

cout << endl;

cout << "Testing the constructor with one integer argument..." << endl;

cout << "Enter a month number: ";

cin >> monthNum;

Month testMonth1(monthNum);

testMonth1.outputMonth_num();

testMonth1.outputMonthLetters();

//

// Construct a month using the constructor with three letters as arguments

//

cout << endl;

cout << "Testing the constructor with 3 letters as arguments ..." << endl;

cout << "Enter the first three letters of a month (lowercase): ";

cin >> firstLetter >> secondLetter >> thirdLetter;

cout << endl;

Month testMonth2(firstLetter, secondLetter, thirdLetter);

testMonth2.outputMonth_num();

testMonth2.outputMonthLetters();

//

// See if user wants to try another month

//

cout << endl;

cout << "Do you want to test again? (y or n) ";

cin >> testAgain;

}

while (testAgain == 'y' || testAgain == 'Y');

return 0;

}

Month::Month(char firstLetter, char secondLetter, char thirdLetter)

{

if ((firstLetter == 'j')&&(secondLetter == 'a')&&(thirdLetter == 'n'))

outputMonth_num = 1;

if ((firstLetter == 'f')&&(secondLetter == 'e')&&(thirdLetter == 'b'))

outputMonth_num = 2;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'r'))

outputMonth_num = 3;

if ((firstLetter = 'a')&&(secondLetter == 'p')&&(thirdLetter == 'r'))

outputMonth_num = 4;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'y'))

outputMonth_num = 5;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(thirdLetter == 'n'))

outputMonth_num = 6;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(.thirdLetter == 'l'))

outputMonth_num = 7;

if ((firstLetter == 'a')&&(secondLetter == 'u')&&(thirdLetter == 'g'))

outputMonth_num = 8;

if ((firstLetter == 's')&&(secondLetter == 'e')&&(thirdLetter == 'p'))

outputMonth_num = 9;

if ((firstLetter == 'o')&&(secondLetter == 'c')&&(thirdLetter == 't'))

outputMonth_num = 10;

if ((firstLetter == 'n')&&(secondLetter == 'o')&&(thirdLetter == 'v'))

outputMonth_num = 11;

if ((firstLetter == 'd')&&(secondLetter == 'e')&&(thirdLetter == 'c'))

outputMonth_num = 12;

}

Month::inputMonthByNumber

{

if (Month_num > 12 && Month_num < 1)

cout << "Invalid number for Month, please choose 1-12)\n";

}

void Month::outputMonth_num()

{

if (month >= 1 && month <= 12)

cout ><< "Month: " << month << endl;

else

cout << "Error - The month is not a valid!" << endl;

}

void Month::outputMonthLetters()

{

switch (month)

{

case 1:

cout << "Jan" << endl;

break;

case 2:

cout << "Feb" << endl;

break;

case 3:

cout << "Mar" << endl;

break;

case 4:

cout << "Apr" << endl;

break;

case 5:

cout << "May" << endl;

break;

case 6:

cout << "Jun" << endl;

break;

case 7:

cout << "Jul" << endl;

break;

case 8:

cout << "Aug" << endl;

break;

case 9:

cout << "Sep" << endl;

break;

case 10:

cout << "Oct" << endl;

break;

case 11:

cout << "Nov" << endl;

break;

case 12:

cout << "Dec" << endl;

break;

default:

cout << "Error - the month is not a valid!" << endl;

}

}

7 0

3 years ago

A well-insulated rigid vessel contains 3 kg of saturated liquid water at 40oC. The vessel also contains an electrical resistor t

user100 [1]

Answer:

The final temperature is 111.66°C

Explanation:

The given conditions :-

i) Well insulated means no heat loss.

ii) Rigid vessels means volume remains same.

iii) Initial temperature ( T₁ ) = 40°C. = 273 + 40 = 313 K

iv ) Mass of water in vessel = 3 kg.

v) current drawn by resistor ( i ) = 10 ampere.

vi) Voltage applied ( V ) = 50 volts.

vii) The time for which resistor operating ( t ) = 30 minute = 30 * 60 = 1800 seconds.

Now we have to calculate heat developed by resistor in vessel.

Q = V * i * t = 50 * 10 * 1800 = 900,000 J = 900 KJ.

Since it is a rigid container so the work done is zero.

Q = du ( du - change in internal energy)

Q = m * C * dT ( C = 4.186 KJ/KgK )

Q = 3 * 4.186 * (T₂ - T₁ )

900 = 12.558 * ( T₂ - 313 )

T₂ - 313 = 71.6674

T₂ = 384.6674 K

T = 384.6674 - 273 = 111.66°C

So the final temperature is 111.66°C.

3 0

3 years ago