Answer:
2.9*10^14 electrons
Explanation:
An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.
An electron has a charge of 1.6*10^-19 C.
35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons
So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.
Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
Q = 5.06 x 10⁻⁸ m³/s
Answer:
#include <iostream>
#include <string>
#include "user.h"
#include "password.h"
using namespace Authenticate;
using namespace std;
int main()
{
inputUserName();
inputPassword();
cout << "Your username is " << getUserName() <<
" and your password is: " <<
getPassword() << endl;
return 0;
}
user.h:
#ifndef USER_H
#define USER_H
#include <string>
using namespace std;
namespace Authenticate
{
namespace
{
bool isvalid();
}
void inputUserName();
string getUserName();
}
#endif
user.cpp:
#include <iostream>
#include "user.h"
namespace Authenticate
{
string username="";
namespace
{
bool isvalid()
{
if(username.length() == 8)
return true;
else
return false;
}
}
void inputUserName(){
do
{
cout << "Enter your username (8 letters only)" << endl;
cin >> username;
}
while(!isvalid());
}
string getUserName()
{
return username;
}
}
password.h:
#ifndef PASSWORD_h
#define PASSWORD_h
#include <string>
using namespace std;
namespace Authenticate
{
namespace
{
bool isValid();
}
void inputPassword();
string getPassword();
}
#endif
password.cpp:
#include <iostream>
#include <string>
using namespace std;
namespace Authenticate
{
string password="";
namespace
{
bool isValid()
{
if(password.length() >= 8)
{
for(int i=0; i<password.length(); i++)
if(password[i] >= '0' && password[i] <= '9')
return true;
return false;
}
else
return false;
}
}
void inputPassword(){
do
{
cout << "Enter your password (at least 8 characters " <<
"and at leat one non-letter)" << endl;
cin >> password;
}
while(!isValid());
}
string getPassword()
{
return password;
}
}
Answer:
the magnitude of F_A is 752 N
the direction theta of F_A is 57.9°
Explanations:
Given that,
Resultant force = 1330 N in x direction
∑Fx = R
from the diagram of the question which i uploaded along with this answer
FB = 800 N
FAsin∅ + FBcos30 = 1330 N
FAsin∅ = 1330 - (800 × cos30)
FA = 637.18 / sin∅
Now ∑Fx = 0
FAcos∅ - FBsin30 = 0
we substitute for FA
(637.18 / sin∅)cos∅ = 800 × sin30
637.18 / 800 × sin30 = sin∅/cos∅
and we know that { sin∅/cos∅ = tan∅)
so tan∅ = 1.59295
∅ = 57.88° ≈ 57.9°
THEREFORE FROM THE EQUATION
FA = 637.18 / sin∅
we substitute ∅
so FA = 637.18 / sin57.88
FA = 752 N
