Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
c) By using the equation of the ideal gases you obtain:
Answer:
The correct answer is C. World Coordinate System
Explanation:
The World Coordinate System has to do with that coordinate system which is fixed in the activities of the CADing. There is a default system in which we can refer to them as soon as we want to manipulate the objects and add new elements.
The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000 
(40 10^-3) / π (5 10^-3)^3
σ = 305 10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305 10^6) (15 10^-3)^3 / 3(40 10^-3)
F = 17156 N.
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:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
