The answer to this problem is the results of the point A
Answer:
D.
Explanation:
In automated welding, defined as “welding with equipment that requires only occasional or no observation of the weld, and no manual adjustment of the equipment controls,” the welder's involvement is limited to activating the machine to initiate the welding cycle and observing the weld on an intermittent basis.
Answer:
23.3808 kW
20.7088 kW
Explanation:
ρ = Density of oil = 800 kg/m³
P₁ = Initial Pressure = 0.6 bar
P₂ = Final Pressure = 1.4 bar
Q = Volumetric flow rate = 0.2 m³/s
A₁ = Area of inlet = 0.06 m²
A₂ = Area of outlet = 0.03 m²
Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s
Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s
Height between inlet and outlet = z₂ - z₁ = 3m
Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation
Work done by pump
∴ Power input to the pump 23.3808 kW
Now neglecting kinetic energy
Work done by pump
∴ Power input to the pump 20.7088 kW
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.
