Answer:
h = 287.1 m
Explanation:
the density of mercury \rho =13570 kg/m3
the atmospheric pressure at the top of the building is

the atmospheric pressure at bottom


we have also

1.18*9.81*h = (100.4 -97.08)*10^3
h = 287.1 m
Explanation:
The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.
Answer:
It would take approximately 305 s to go to 99% completion
Explanation:
Given that:
y = 50% = 0.5
n = 1.7
t = 100 s
We need to first find the parameter k from the equation below.

taking the natural logarithm of both sides:

Substituting values:

Also
![t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}](https://tex.z-dn.net/?f=t%5En%3D-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%5C%5Ct%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D)
Substituting values and y = 99% = 0.99
![t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s](https://tex.z-dn.net/?f=t%3D%5Csqrt%5Bn%5D%7B-%5Cfrac%7Bln%281-y%29%7D%7Bk%7D%7D%3D%5Csqrt%5B1.7%5D%7B-%5Cfrac%7Bln%281-0.99%29%7D%7B2.76%2A10%5E%7B-4%7D%7D%7D%3D304.6s)
∴ t ≅ 305 s
It would take approximately 305 s to go to 99% completion
Answer:
The following program is in C++.
#include <bits/stdc++.h>
using namespace std;
void lastChars(string s)
{
int l=s.length();
if(l!=0)
{
cout<<"The last character of the string is: "<<s[l-1];
}
}
int main() {
string s;//declaring a string..
getline(cin,s);//taking input of the string..
lastChars(s);//calling the function..
return 0;
}
Input:-
Alex is going home
Output:-
The last character of the string is: e
Explanation:
In the function lastChars() there is one argument that is a string.I have declared a integer variable l that stores the length of the string.If the length of the string is not 0.Then printing the last character of the string.In the main function I have called the function lastChars() with the string s that is prompted from the user.