Answer:
// Program is written in C++
// Comments are used to explain some lines
// Only the required function is written. The main method is excluded.
#include<bits/stdc++.h>
#include<iostream>
using namespace std;
int divSum(int num)
{
// The next line declares the final result of summation of divisors. The variable declared is also
//initialised to 0
int result = 0;
// find all numbers which divide 'num'
for (int i=2; i<=(num/2); i++)
{
// if 'i' is divisor of 'num'
if (num%i==0)
{
if (i==(num/i))
result += i; //add divisor to result
else
result += (i + num/i); //add divisor to result
}
}
cout<<result+1;
}
C. Route and roadways defined as class I highways
Answer:
D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted
Explanation:
access-list 90
deny 10.168.7.0 0.0.0.255
permit 10.168.7.10
permit 10.168.7.11
permit 10.168.7.12
deny any
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
