Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.
<u>Given the following data:</u>
- Jam density = 122 veh/km.
<h3>How to calculate the
maximum flow.</h3>
According to Greenshield Model, maximum flow is given by this formula:
<u>Where:</u>
is the free flow speed.
is the Jam density.
In order to calculate the free flow speed, we would use this formula:
Substituting the parameters into the model, we have:
Max flow = 3,599 veh/hr.
Read more on traffic flow here: brainly.com/question/15236911
Answer:
//Define the header file
#ifndef PLAYER_H
#define PLAYER_H
//header file.
#include <string>
//Use the standard namespace.
using namespace std;
//Define the class Player.
class Player
{
//Declare the required data members.
string name;
int score;
public:
//Declare the required
//member functions.
void setName(string par_name);
void setScore(int par_score);
string getName();
int getScore();
}
//End the definition
//of the header file.
#endif
Player.cpp:
//Include the "Player.h" header file,
#include "Player.h"
//Define the setName() function.
void Player::setName(string par_name)
{
name = par_name;
}
//Define the setScore() function.
void Player::setScore(int par_score)
{
score = par_score;
}
//Define the getName() function.
string Player::getName()
{
return name;
}
//Define the getScore() function.
int Player::getScore()
{
return score;
}
Some casting alloys can be heat-treated after casting for added strength, True
