A set of bus bars also can be called a

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(n)_____ is a device that provides the power and motion to manipulate the moving parts of a valve or damper used to control flu

Lesechka [4]

Answer:

Out of the four options provided

option A. actuator

is correct

Explanation:

An actuator is the only device out of the four mentioned devices that provides power and ensures the motion in it in order to manipulate the movement of the moving parts of the damper or a valve used whereas others like ratio regulator are used to regulate air or gas ratio and none mof the 3 remaining options serves the purpose

5 0

3 years ago

A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

Density = 61 veh/km.

Speed = 59 km/hr.

Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

$q_{max}=\frac{V_f \times K_i}{4}$

<u>Where:</u>

$V_f$ is the free flow speed.

$K_i$ is the Jam density.

In order to calculate the free flow speed, we would use this formula:

$V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr$

Substituting the parameters into the model, we have:

$q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}$

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0

2 years ago

Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper

Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

$\frac{w}{Q1} = \frac{35}{100}$

W = $1.2*\frac{35}{100}*1000kj/s$

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

$\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ$

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0

2 years ago

Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)

AveGali [126]

The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

Since $V_{ds}$ > $V_{gs} - Vt$ because $V_{gs}$ > Vt.
By the saturation region the MOSFET is operating.
A specific source voltage and gate of NMOS, the voltage get drained during the specific level, the drain voltage is rises beyond where there is no effect of current during saturated region.
MOSFET is a transistor which is a device of semiconductor vastly used for the electronic amplifying signals and switching in the devices of electronics.
The core of this is integrated circuit.
It is fabricated and designed in an individual chips due to tiny sizes.

7 0

3 years ago