Answer:
Relational query language is the language that is used in the queries of the relational databases that has rows and columns in table. The user or client presents a request for gaining the information from the database. The relationships in the database are defined in numerous ways which creates the query.
This language can be procedural form or non-procedural form.The operations performed by this language are communication with the relational database, analyzing the relationships between the entities of database,splitting the request from client and then execution of that request is done by database management system(DBMS), etc.
Answer:
C++ code is given below
Explanation:
#include <iostream>
#include <cctype>
#include <string.h>
#include <cstring>
#include <sstream>
using namespace std;
struct Car {
public:
char reportingMark[5];
int carNumber;
string kind;
bool loaded;
string destination;
};
void input(Car *);
void output(Car *);
int main() {
Car *T = new Car;
input(T);
output(T);
delete T;
return 0;
}
void input(Car *T)
{
string str, s;
cout << " Enter the reporting mark as a 5 or less character uppercase string: ";
cin >> str;
for (int i = 0; i < str.length(); i++)
T->reportingMark[i] = toupper(str[i]);
cout << " Enter the car number: ";
cin >> T->carNumber;
cout << " Enter the kind: ";
cin >> T->kind;
cout << " Enter the loaded status as true or false: ";
cin >> s;
istringstream(s) >> boolalpha >> T->loaded;
if (T->loaded == true) {
cout << " Enter the destination: ";
cin.ignore();
getline(cin, T->destination);
}
else
T->destination = "NONE";
}
void output(Car *T)
{
cout << " Reporting Mark: " << T->reportingMark;
cout << " Car Number: " << T->carNumber;
cout << " Kind: " << T->kind;
cout << " Loaded Status: " << boolalpha << T->loaded;
cout << " Destination: " << T->destination << " ";
}
The item that would be most likely to keep in a database would be the payroll records. The other items such as address book, financial statements and sales reports would not necessarily be placed in a database.
The answer would be letter A.
Answer:
t= 8.7*10⁻⁴ sec.
Explanation:
If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.
As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.
Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:
If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:
⇒ t = 8.7*10⁻⁴ sec.
