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omeli [17]
3 years ago
11

A transfer function has the following zeros and poles: one zero at s = 0 and one zero at s = [infinity]; one pole at s = −100 an

d one pole at s = −106. The magnitude of the transfer function at ω = 104 rad/s is 100. Find the transfer function T(s) and sketch a Bode plot for its magnitude.
Engineering
1 answer:
const2013 [10]3 years ago
6 0

harden you could either me or leave

harden you could either me or leave

harden you could either me or leave

v

harden you could either me or leave

vvvv

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Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67
defon

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;

}

5 0
3 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

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Explanation:

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