Answer:
The time necessary to purge 95% of the NaOH is 0.38 h
Explanation:
Given:
vfpure water(i) = 3 m³/h
vNaOH = 4 m³
xNaOH = 0.2
vfpure water(f) = 2 m³/h
pwater = 1000 kg/m³
pNaOH = 1220 kg/m³
The mass flow rate of the water is = 3 * 1000 = 3000 kg/h
The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg
When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg
The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³
The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
Answer:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main ()
{
// Variable declaration
string name;
int numQuestions;
int numCorrect;
double percentage;
//Prompt to enter student's first and last name
cout<<"Enter student's first and last name";
cin>>name; // this line accepts input for variable name
cout<<"Number of question on test"; //Prompt to enter number of questions on test
cin>> numQuestions; //This line accepts Input for Variable numQuestions
cout<<"Number of answers student got correct: "; // Prompt to enter number of correct answers
cin>>numCorrect; //Enter number of correct answers
percentage = numCorrect * 100 / numQuestions; // calculate percentage
cout<<name<<" "<<percentage<<"%"; // print
return 0;
}
Explanation:
The code above calculates the percentage of a student's score in a certain test.
The code is extracted from the Question and completed after extraction.
It's written in C++ programming language
Answer:
Hydrostatic force = 41168 N
Explanation:
Complete question
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Let "x" be the side length submerged in water.
Then
w(x)/base = (4+3-x)/altitude
w(x)/5 = (4+3-x)/3
w(x) = 5* (7-x)/3
Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7
HF = integration of 40x - 4x^2/3
HF = 20x^2 - 4x^3/9 with limit 4 to 7
HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))
HF = 658.69 N *62.5 = 41168 N
Answer:
Side effects - sudden loss of balance/ repeated falls
Outputs - sever sickness and could me factual
Inputs/corrections of this- medications and experimental treatments to help slow the process of deterioration
Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = 
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ = 
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
