Explicar el funcionamiento de un multímetro analógico.
1 answer:
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Answer:
Explanation:
attached is an uploaded picture to support the answer.
the program is written thus;
#include<iostream>
using namespace std;
// function declaration
void rShift(int&, int&, int&, int&, int&, double&);
int main()
{
// declare the variables
int a1, a2, a3, a4;
int maximum = 0;
double average = 0;
// inputting the numbers
cout << "Enter all the 4 integers seperated by space -> ";
cin >> a1 >> a2 >> a3 >> a4;
cout << endl << "Value before shift." << endl;
cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "
<< a3 << ", a4 = " << a4 << endl << endl;
// calling rSift()
// passing the actual parameters
rShift(a1,a2,a3,a4,maximum,average);
// printing the values
cout << "Value after shift." << endl;
cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "
<< a3 << ", a4 = " << a4 << endl << endl;
cout << "Maximum value is: " << maximum << endl;
cout << "Average is: " << average << endl;
}
// function to right shift the parameters circularly
// and calculate max, average of the numbers
// and return it to main using pass by reference
void rShift(int &n1, int &n2, int &n3, int &n4, int &max, double &avg)
{
// calculating the max
max = n1;
if(n2 > max)
max = n2;
if(n3 > max)
max = n3;
if(n4 > max)
max = n4;
// calculating the average
avg = (double)(n1+n2+n3+n4)/4;
// right shifting the numbers circulary
int temp = n2;
n2 = n1;
n1 = n4;
n4 = n3;
n3 = temp;
}
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.