Answer:
#include<iostream>
int main() {
float num_1, num_2, num_3, num_4, average;
//Taking input for four numbers
std::cout << "Enter first number(integer or floating-point)";
std::cin >> num_1;
std::cout << "Enter second number(integer or floating-point)";
std::cin >> num_2;
std::cout << "Enter third number(integer or floating-point)";
std::cin >> num_3;
std::cout << "Enter fourth number(integer or floating-point)";
std::cin >> num_4;
average= (num_1+num_2+num_3)/3;
// Comparing average with fourth number
if (average==num_4)
{
std::cout << "Equal";
}
return 0;
}
Answer:
- #include <iostream>
- using namespace std;
- void printLarger(int a, int b){
-
- if(a > b){
- cout<<a;
- }else{
- cout<<b;
- }
- }
- int main()
- {
- printLarger(4, 5);
- return 0;
- }
Explanation:
The solution code is written in C++.
Firstly define a function printLarger that has two parameters, a and b with both of them are integer type (Line 5). In the function, create an if condition to check if a bigger than b, print a to terminal (Line 7-8). Otherwise print b (Line 9-10).
In the main program, test the function by passing 4 and 5 as arguments (Line 16) and we shall get 5 printed.
Answer:
Harmful effect associated with extraction of Aluminum, Gold and Copper are:
During the melting of aluminium there is a released of per fluorocarbon are more harmful than carbon dioxide in the environment as they increased the level of green house gases and cause global warming. The process of transforming raw material into the aluminium are much energy intensive.
Gold mining industries destroyed land scopes and increased the amount of toxic level in the environment and they also dump there toxic waste in the natural water bodies, which increased the level of water pollution in the environment.
Copper mining causes the health problems like asthma and problem in respiratory system because of the inhalation of silica dust. It also increased the level of sulfur diode in the environment which cause acid rain and destroyed various trees and buildings in the nature.
Answer:
L = 46.35 m
Explanation:
GIVEN DATA
\dot m = 0.25 kg/s
D = 40 mm
P_1 = 690 kPa
P_2 = 650 kPa
T_1 = 40° = 313 K
head loss equation
![[\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m](https://tex.z-dn.net/?f=%5B%5Cfrac%7BP_1%7D%7B%5Crho%7D%20%2B%5Calpha%20%5Cfrac%7Bv_1%5E2%7D%7B2%7D%20%2Bgz_1%5D%20-%5B%5Cfrac%7BP_2%7D%7B%5Crho%7D%20%2B%5Calpha%20%5Cfrac%7Bv_2%5E2%7D%7B2%7D%20%2Bgz_2%5D%20%3D%20h_l%20%2Bh_m)
where
density is constant
head is same so,
curvature is constant so
neglecting minor losses
we know
is given as
therefore
V = 25.90 m/s
for T = 40 Degree, 
Re = 4.16*10^5 > 2300 therefore turbulent flow
for Re =4.16*10^5 , f = 0.0134
Therefore
L = 46.35 m
