The type of documentation that Holly should use is to generate a graphic representation of the Apple system.
<h3>What is Reverse Engineering?</h3>
This refers to the methodical act of dismantling an object in order to learn how it works or was constructed.
Hence, the use of a graphic representation of the Apple system which Holly wants to reverse engineer is the best option because it uses a standard document UML notation.
Read more about reverse engineering here:
brainly.com/question/25682883
Answer:
Vx = 6.242 x 10raised to power 15
Vy = -6.242 x 10raised to power 15
Explanation:
from E = IVt
but V = IR from ohm's law and Q = It from faraday's first law
I = Q/t
E = Q/t x V x t = QV
hence, E =QV
V = E/Q
BRIGHT HEADLIGHTS
AND SEVERE WEATHER CONDITIONS
Answer:
#include <iostream>
#include <string>
using namespace std;
bool isPalindrome(string str)
{
int length = str.length();
for (int i = 0; i < length / 2; i++)
{
if (tolower(str[i]) != tolower(str[length - 1 - i]))
return false;
}
return true;
}
int main()
{
string s[6] = {"madam", "abba", "22", "67876", "444244", "trymeuemyrt"};
int i;
for(i=0; i<6; i++)
{
//Testing function
if(isPalindrome(s[i]))
{
cout << "\n " << s[i] << " is a palindrome... \n";
}
else
{
cout << "\n " << s[i] << " is not a palindrome... \n";
}
}
return 0;
}
Answer:
23.3808 kW
20.7088 kW
Explanation:
ρ = Density of oil = 800 kg/m³
P₁ = Initial Pressure = 0.6 bar
P₂ = Final Pressure = 1.4 bar
Q = Volumetric flow rate = 0.2 m³/s
A₁ = Area of inlet = 0.06 m²
A₂ = Area of outlet = 0.03 m²
Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s
Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s
Height between inlet and outlet = z₂ - z₁ = 3m
Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation
![\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m](https://tex.z-dn.net/?f=%5Cfrac%20%7BP_1%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_1%5E2%7D%7B2g%7D%2Bz_1%2Bh%3D%5Cfrac%20%7BP_2%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_2%5E2%7D%7B2g%7D%2Bz_2%5C%5C%5CRightarrow%20h%3D%5Cfrac%7BP_2-P_1%7D%7B%5Crho%20g%7D%2B%5Cfrac%7BV_2%5E2-V_1%5E2%7D%7B2g%7D%2Bz_2-z_1%5C%5C%5CRightarrow%20h%3D%5Cfrac%7B%281.4-0.6%29%5Ctimes%2010%5E5%7D%7B800%5Ctimes%209.81%7D%2B%5Cfrac%7B6.67_2%5E2-3.33%5E2%7D%7B2%5Ctimes%209.81%7D%2B3%5C%5C%5CRightarrow%20h%3D14.896%5C%20m)
Work done by pump
![W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W](https://tex.z-dn.net/?f=W_%7Bp%7D%3D%5Crho%20gQh%5C%5C%5CRightarrow%20W_%7Bp%7D%3D800%5Ctimes%209.81%5Ctimes%200.2%5Ctimes%2014.896%5C%5C%5CRightarrow%20W_%7Bp%7D%3D23380.8%5C%20W)
∴ Power input to the pump 23.3808 kW
Now neglecting kinetic energy
![h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\](https://tex.z-dn.net/?f=h%3D%5Cfrac%7BP_2-P_1%7D%7B%5Crho%20g%7D%2Bz_2-z_1%5C%5C%5CRigharrow%20h%3D%5Cfrac%7B%281.4-0.6%29%5Ctimes%2010%5E5%7D%7B800%5Ctimes%209.81%7D%2B3%5C%5C%5CRigharrow%20h%3D13.19%5C%20m%5C%5C)
Work done by pump
![W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W](https://tex.z-dn.net/?f=W_%7Bp%7D%3D%5Crho%20gQh%5C%5C%5CRightarrow%20W_%7Bp%7D%3D800%5Ctimes%209.81%5Ctimes%200.2%5Ctimes%2013.193%5C%5C%5CRightarrow%20W_%7Bp%7D%3D20708.8%5C%20W)
∴ Power input to the pump 20.7088 kW