Base on the question, and in my further computation, the possible answers would be the following and I hope you are satisfied with my answer and feel free to ask for more.
- If you want to determine the Thevenin equivalent voltage and resistance without overloading the battery, then apply some known resistance
<span><span>RL</span><span>RL</span></span> and measure the output voltage as <span><span>VL</span><span>VL</span></span>. Measure the voltage without a load as <span><span>V<span>OC</span></span><span>V<span>OC</span></span></span>. The voltage divider equation tells us that
<span><span><span>VL</span>=<span>V<span>OC</span></span><span><span>RL</span><span><span>R<span>TH</span></span>×<span>RL</span></span></span></span><span><span>VL</span>=<span>V<span>OC</span></span><span><span>RL</span><span><span>R<span>TH</span></span>×<span>RL</span></span></span></span></span>
Solve for <span><span>R<span>TH</span></span><span>R<span>TH</span></span></span>, and you know that <span><span><span>V<span>TH</span></span>=<span>V<span>OC</span></span></span><span><span>V<span>TH</span></span>=<span>V<span>OC</span></span></span></span>.
Answer:
-26
Explanation:
The given binary number is 1110 0101. Also given that the signed binary number is represented using one's compliment.
We begin by computing the 1s complement representation of 1110 0101 by inverting the bits: 00011010
Converting 00011010 to decimal, it corresponds to 26.
So the 1s complement of the original number is 26. This means that the original number was -26.
Spring break around Easter I would believe
Answer:
C++ code is given below
Explanation:
#include<iostream>
#include <cstring>
using namespace std;
int housekeeping(string EOFNAME);
int mainLoop(string name,string EOFNAME);
int finish();
void main()
{
string name;
string EOFNAME = "ZZZZ";
cout << "enter the name" << endl;
cin >> name;
if (name != EOFNAME)
{
housekeeping(EOFNAME);
}
if (name != EOFNAME)
{
mainLoop(name , EOFNAME);
}
if (name != EOFNAME)
{
finish();
}
system("pause");
}
int housekeeping(string EOFNAME)
{
cout << "enter first name " << EOFNAME << " to quit " << endl;
return 0;
}
int mainLoop(string name, string EOFNAME)
{
int hours;
int rate,gross;
int DEDUCTION = 45;
int net;
cout << "enter hours worked for " << name << endl;
cin >> hours;
cout << "enter hourly rate for " << name << endl;
cin >> rate;
gross = hours*rate;
net = gross - DEDUCTION;
if (net > 0)
{
cout << "net pay for " << name << " is " << net << endl;
}
else
{
cout << "dedections not covered.net is 0.";
}
cout << "enter next name or " << EOFNAME << " to quit" << endl;
cin >> name;
return 0;
}
int finish()
{
cout << "end of job"<<endl;
return 0;
}
Answer:
The algorithm is as follows;
1. Start
2. Input TeddyBears
3. Input Hours
4. WagebyTeddy = 2 * TeddyBears
5. WagebyHour = 5 * Hours
6. If WagebyHour > WagebyTeddy then
6.1 Print WagebyHour
7. Else
7.1. Print WagebyTeddy
8. Stop
Explanation:
The following variables are used;
TeddyBears -> Number of teddy bears made
Hours -> Number of Hours worked
WagebyTeddy -> Wages for the number of teddy bears made
WagebyHour -> Wages for the number of hours worked
The algorithm starts by accepting input for the number of teddy bears and hours worked from the user on line 2 and line 3
The wages for the number of teddy bears made is calculated on line 4
The wages for the number of hours worked is calculated on line 5
Line 6 checks if wages for the number of hours is greated than wages for the number of bears made;
If yes, the calculated wages by hour is displayed
Otherwise
the calculated wages by teddy bears made is displayed
