Answer:
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
int main() {
string name[5];
int age[5];
int i,j;
for ( i = 0; i<=4; i++ ) {
cout << "Please enter student's name:";
cin >> name[i];
cout << "Please enter student's age:";
cin >> age[i];
}
for (i=0;i<=4;i++){
cout<<"Age of "<< name[i]<<" is "<<age[i]<<endl;
}
}
Output of above program is displayed in figure attached.
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
Answer:
what are simple machines?
Explanation:
it is 2020 let's be honest all
The efficiency of a transformer is mainly dependent on a)- Core losses
Hope this helps! :)
Answer:
a)
, b)
Explanation:
a) The Reynolds number for the water flowing in a circular tube is:

Let assume that density and dynamic viscosity at 25 °C are
, respectively. Then:


b) The result is:
