Answer:
Following are the solution to this question:
Explanation:
The number of vacancies by the cubic meter is determined.
Answer: a) 0.948 b) 117.5µf
Explanation:
Given the load, a total of 2.4kw and 0.8pf
V= 120V, 60 Hz
P= 2.4 kw, cos θ= 80
P= S sin θ - (p/cos θ) sin θ
= P tan θ(cos^-1 (0.8)
=2.4 tan(36.87)= 1.8KVAR
S= 2.4 + j1. 8KVA
1 load absorbs 1.5 kW at 0.707 pf lagging
P= 1.5 kW, cos θ= 0.707 and θ=45 degree
Q= Ptan θ= tan 45°
Q=P=1.5kw
S1= 1.5 +1.5j KVA
S1 + S2= S
2.4+j1.8= 1.5+1.5j + S2
S2= 0.9 + 0.3j KVA
S2= 0.949= 18.43 °
Pf= cos(18.43°) = 0.948
b.) pf to 0.9, a capacitor is needed.
Pf = 0.9
Cos θ= 0.9
θ= 25.84 °
(WC) V^2= P (tan θ1 - tan θ2)
C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2
f=60, π=22/7
C= 117.5µf
Answer:
Yes. YES yes yes. Unless you are in Australia or something.
Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime(int bagOunces) {
if(bagOunces < 3){
cout << "Too small";
cout << endl;
}
else if(bagOunces > 10){
cout << "Too large";
cout << endl;
}
else{
cout << (6 * bagOunces) << " seconds" << endl;
}
}
int main() {
PrintPopcornTime(7);
return 0;
}
Explanation:
Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.
