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3 years ago

Resistance depends on which three properties of a wire?

Engineering

1 answer:

kenny6666 [7]3 years ago

7 0

Answer:

Explanation:

From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.

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Which of the following allows team members to visualize a design model from a variety of perspectives?

julsineya [31]

Answer: from what i know im pretty sure its isometrics or sketches im certain its sketches but not 100%

Explanation: A sketch is a rapidly executed freehand drawing that is not usually intended as a finished work. A sketch may serve a number of purposes: it might record something that the artist sees, it might record

8 0

3 years ago

Read 2 more answers

What is the connection between the air fuel ratio and an engine running rich/poor? please give clear examples and full sentances

gavmur [86]

Explanation:

Air fuel ratio:

Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that

$Air\ fuel\ ratio=\dfrac{mass\ of\ air}{mass\ of\ fuel}$

As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.

When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.

When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.

5 0

3 years ago

When asked about favorite Thanksgiving leftovers, 9% of the people said turkey and 7100 said mashed potatoes. Which food is more

irakobra [83]

Answer:

how many people were asked though

Explanation:

8 0

2 years ago

Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is

ddd [48]

Answer:

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

cout << "\n----------------------\n";

for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

int firstDay = day-1;

//Space print

for(int i=1; i< firstDay; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

int cellCnt = 0;

// Iteration of days

for(int i=1; i<=numDays; i++)

{

//Output days

cout << left << setw(1) << "|" << setw(2) << i;

cellCnt += 1;

// New line

if ((i + firstDay-1) % 7 == 0)

{

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

cellCnt = 0;

}

// Empty cell print

if (cellCnt != 0)

{

// For printing spaces

for(int i=1; i<7-cellCnt+2; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

cout << "\n----------------------\n";

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end

3 0

3 years ago

A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The

AleksandrR [38]

Answer:

The work furnished by the compressor is $69.77kJ/s$

The minimum work required for the state to change is $55.26kW$

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

8 0

3 years ago