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Tema [17]
2 years ago
5

2. How many amperes of current will flow through a circuit that has 3 ohms of resistance if

Engineering
1 answer:
Dmitriy789 [7]2 years ago
4 0

Answer:

4A

Explanation:

From Ohms law ;

R= V/I where R is resistance, V is voltage and I is current

Given that;

R= 3Ω

V= 12

Using the values in the formula as;

R= V/I

3=12/I

I= 12/3

I= 4 A

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Lately, you have noticed some repetitive stress in your wrist. Which sign is most likely the cause of that stress and pain?
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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;

}

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2 years ago
Select the correct text in the passage.
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3 years ago
Read 2 more answers
Determine the following parameters for the water having quality x=0.7 at 200 kPa:
ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

   Temperature = 120.21°C

c). Specific volume

  $v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

  $v_x=v_f+x(v_g-v_f)$

       $=0.001061+0.7(0.88578-0.001061)$

       $=0.62036 \ m^3/kg$

d). Specific energy (u_x)

    $u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

   $u_x=504.5 + 0.7(2024.6)$

         $=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

   At $h_f = 504.71, \ \ h_{fg} = 2201.6$

   h_x=504.71+(0.7\times 2201.6)

        $= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

  $H=mh_x$

       $= 0.5 \times 2045.83$

       = 1022.91 kJ

7 0
3 years ago
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