Answer:
The correct answer to the following question is option A. an avoidance spiral.
Explanation:
An avoidance spiral is a spiral in which parties avoid the conflicts at all cost in which leads to reducing the influence, dependence, etc. This spiral can also be destructive.
In other words, it is the only spiral in parties slowly reduces their dependence and the influence on the one and another.
Answer:
The Java class is given below with appropriate tags for better understanding
Explanation:
public class Cat extends Pet{
private String breed;
public Cat(String name, String owner, String breed){
/* implementation not shown */
super(name, owner);
this.breed = breed;
}
public String getBreed() {
return breed;
}
public void setBreed(String breed) {
this.breed = breed;
}
public String speak(){ /* implementation not shown */
return "Purring…";
}
}
Answer:
Explanation:
Given
Power at point A = 100W
Power at point B = 90W
Required
Determine the attenuation in decibels
Attenuation is calculated using the following formula
Where 
t and 
t
Substitute these values in the given formula
<em>(Approximated)</em>
Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3
