Given that,
Resistance, R = 30 ohms
Current, I = 10 A
To find,
The voltage of the battery.
Solution,
Let V is the voltage of the battery. We can use Ohm's law to find V.
Since, V = IR
Put all the values in the above formula.
V = 10×30
V = 300 Volt
So, the voltage of the battery in the voltmeter is 300 Volt.
Answer :
Written in java
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
int year;
Scanner scanner = new Scanner(System.in);
System.out.print("Please enter a year\n\n");
year = scanner.nextInt();
while (year < 1582) {
System.out.print("\nPlease enter a different year above 1582\n");
year = scanner.nextInt();
}
if (year % 4 == 0) {
if(year % 100 ==0 && year % 400 != 0){
System.out.println(year + " is not a leap year\n\n");
}else {
System.out.println(year + " is a leap year\n");
}
} else {
System.out.println(year + " is not a leap year\n\n");
}
}
}
Try resending it isn't there a button to resend the email? and look in spam
Answer:
Early in the history of technology, the development of tools and machines was based on technical know-how versus scientific knowledge as is done today.
Explanation:
In the era prior to technological development, men had a basic and at times rudimentary manufacturing development, based on artisan processes that developed their tools and machinery by hand, without any type of automation or mass production rules.
Thus, each part or tool was manufactured in a unique way, which required a broad mastery of the production process by the manufacturer. This is how the first professions began to emerge, such as blacksmiths for example, who mastered the technique of manufacturing different implements and carried them out without any scientific knowledge.
#include
#include
using namespace std;
int main(){
int input[] = {-19, 34, -54, 65, -1};
std::vector voutput:
std::vector vinput (input, input + sizeof(input) / sizeof(int) );
for (std::vector::iterator it = vinput.begin(); it != vinput.end(); ++it)
if(*it > 0) voutput.insert(voutput.begin(), *it);
for(std::vector::iterator it = voutput.begin(); it < voutput.end(); ++it)
std::cout << *it << ‘\n’ ;
return 0;
}
