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;
}
Answer:
IDK
Explanation:
same thing is happening to me
Answer:
(a) 0 kJ
(b) 9.81 kJ
(c) 31.32 m/s
Explanation:
(a)
From the law of conservation of energy, energy can only be transformed from one state to another. At a height of 50 m, all the kinetic energy is converted to potential energy hence KE=0
(b)
Potential energy, PE=mgh where m is the mass, g is acceleration due to gravity and h is the height
Substituting 50 m for h and 20 Kg for m, taking g as 9.81 then
PE=20*9.81*50=9810 J=9.81 kJ
(c)
Relating the equation of potential energy to the equation of kinetic energy, which is 
where v is the velocity of the mass
Substituting 50 m for h and taking g as 9.81 then
Answer:
The surface temperature of the ground is = 296.946K
Explanation:
Solution
Given
r₁= 0.05m
r₂= 0.08m
Tn =Ti = 77K
Ki = 0.0035 Wm-1K-1
Kg = 1 Wm-1K-1
Z= 2m
Now,
The outer type temperature (Skin temperature pipe)
Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)
Thus,
10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05
⇒ T₀ -77 = 231.72
T₀= 290.72K
The shape factor between the cylinder and he ground
S = 2πL/ln 4z/D
where L = length of pipe
D = outer layer of pipe
S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m
The heat gained in the pipe is = S * Kg * (Tg- T₀)
(10* 1) = 1.606 * 1* (Tg- 290.72)
Tg - 290.72 = 6.2266
Tg = 296.946K
Therefore the surface temperature to the ground is 296.946K
Answer: Technician A, I think
