Two wires of the same length and cross sectional area have finite but different E-field magnitudes inside them when the same current is flowing in each. The E-field magnitude in wire 1 is less than the field magnitude in wire 2. Select all of the following statements that are true based on this situation: The voltage drop in wire 1 is the same as the voltage drop in wire 2
The conductivity of wire 1 is less than the conductivity in wire 2
There is not enough information given in the problem to definitely determine the answer to the other 9 statements.
The power dissipation in wire 1 is more than the power dissipation in wire 2
The conductivity of wire 1 is more than the conductivity in wire 2
The conductivity of wire 1 is the same as the conductivity in wire 2
The power dissipation in wire 1 is less than the power dissipation in wire 2
The power dissipation in wire 1 is the same as the power dissipation in wire 2
The voltage drop in wire 1 is greater than the voltage drop in wire 2
The voltage drop in wire 1 is less than the voltage drop in wire 2

The voltage drop in wire 1 is more than the voltage drop in wire 2.

The conductivity of wire 1 is less than the conductivity of wire 2.

The power dissipation in wire 1 is less than the power dissipation in wire 2.

Explanation:

The electric field magnitude inside the wire is dependent on the current flow. Voltage drop can cause load regulation errors in wiring. The conductor is then used to maintain the regulated flow of current in the wires. In the given scenario the wire  magnitude is greater than wire 1 magnitude. If the magnitude is smaller there will be less conductivity therefore more voltage drops will be observed.

answer:

Explanation:

i hope dis helps

answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

//main function

int main()

{

  // variable

 int n;

 cout<<"enter a number:";

 // read the input

 cin>>n;

 // find the square root of the absolute value

 float square_root=sqrt(abs(n));

 // print the square_root

 cout<<"square root of "<<n<<" is: "<<square_root<<endl;

  return 0;

}

Explanation:

Read a number from user and assign it to "n".Then find its absolute value with abs() function.Then calculate the square root of the absolute value with sqrt() function.Print the square root of the number.

Output:

enter a number:-17

square root of -17 is: 4.12311

                                                         This is easy

                                                     sinserly Natalie 7 grader (this is not that hard)