Is a compass a analog or a digital sensor?

I ran across this symbol in some Electrical wiring documents and I am unaware of what this means. Any help?

Minchanka [31]

Answer:

Opened Push-button Switch (i.e. a PTM Switch)

Explanation:

Tha's just another symbol for a switch, but this one specifies that the switch is a push-button type of switch.

Since it's not touching and completing the line, the state of the switch is initially open.

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

2 years ago

In an air compressor the compression takes place at a constant internal energy and 50KJ of heat are rejected to the cooling wate

lozanna [386]

Answer:

work is 50 kj

Explanation:

Given data

heat (Q) = 50 kj

To find out

work input for the compression stroke per kilogram of air

Solution

we will apply here "first law of thermodynamics" i.e.

The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.

ΔU = Q – W ................1

here ΔU is change in internal energy, Q is heat and W is work done

here U = 0 because air compressor the compression takes place at a constant internal energy in question

so that by equation 1

Q = W

and Q = 50

so work will be 50 kj

8 0

3 years ago

Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0

stealth61 [152]

Answer:

$118_{10}= 0110111$ 2) $-49_{10}=110001_{2}$ 3) $0_{10}=0:16 \Rightarrow 0_{10}=0_{16}$

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

$118_{10}= 0110111$

2) $-49_{10}$

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

$-49_{10}=110001_{2}$

3) $(-0)_{10}$

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

$(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000$

5 0

3 years ago

Implement this C program by defining a structure for each payment. The structure should have at least three members for the inte

Klio2033 [76]

Answer:

#include<stdio.h>

#include<math.h>

void output_amortized(float loan_amount,float intrest_rate,int term_years)

{

int i,j; //Month

int payments; //Number of payments

float loanAmount; //Loan amount

float anIntRate; //Yealy interest Rate

float monIntRate; //Monthly interest rate

float monthPayment; //Monthly payment

float balance; //Balance due

float monthPrinciple; //Monthly principle paid

float monthPaidInt; //Month interest paid

balance=loan_amount;

//Calculations

//Monthly interest rate

monIntRate = ((intrest_rate/(100*12)));

//Monthly payment

payments=term_years;

monthPayment = (loan_amount * monIntRate * (pow(1+monIntRate, payments)/(pow (1+monIntRate, payments)-1)));

monthPaidInt = balance * monIntRate;

//Amount paid to principle

monthPrinciple = monthPayment-monthPaidInt;

//New balance due

balance = balance - monthPrinciple;

printf("\n\nMonthly payment should be :%.2f\n\n",monthPayment);

printf("============================AMORTIZATION SCHEDUAL==========================\n");

printf("#\tPayment\t\tIntrest\t\tPrinciple\t\tBalance\n");

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

{

printf("%d%9c%.2f%9c%.2f%16c%.2f%14c%.2f\n",(i+1),'$',monthPayment,'$',monthPaidInt,'$',monthPrinciple,'$',balance);

monthPaidInt = balance * monIntRate;

//Amount paid to principle

monthPrinciple = monthPayment-monthPaidInt;

//New balance due

balance = balance - monthPrinciple;

}

int main()

{

float principle,rate;

int termYear;

printf("Enter the loan amount: $");

scanf("%f",&principle);

printf("Enter the intrest rate :%");

scanf("%f",&rate);

printf("Enter the loan duration in years: ");

scanf("%d",&termYear);

output_amortized(principle,rate,termYear);

}

Explanation:

see output

6 0

2 years ago