Answer:
endurance length is 236.64 MPa
Explanation:
data given:
d = 37.5 mm
Sut = 760MPa
endurance limit is
Se = 0.5 Sut
= 0.5*760 = 380 MPa
surface factor is
Ka = a*Sut^b
where
Sut is ultimate strength
for AISI 1040 STEEL
a = 4.51, b = -0.265
Ka = 4.51*380^{-0.265}
Ka = 0.93
size factor is given as
Kb =1.29 d^{-0.17}
Kb = 0.669
Se = Sut *Ka*Kb
= 380*0.669*0.93
Se = 236.64 MPa
therefore endurance length is 236.64 MPa
Answer:
The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.
Explanation:
The detailed steps and appropriate calculation is as shown in the attached file.
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:
Q' = 8 KW.h
Q'=28800 KJ
Explanation:
Given that
Heat Q= 4 KW
time ,t = 2 hours
The amount of energy used in KWh given as
Q ' = Q x t
Q' = 4 x 2 KW.h
Q' = 8 KW.h
We know that
1 h = 60 min = 60 x 60 s = 3600 s
We know that W = 1 J/s
The amount of energy used in KJ given as
Q' = 8 x 3600 = 28800 KJ
Therefore
Q' = 8 KW.h
Q'=28800 KJ
