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photoshop1234 [79]
3 years ago
14

How much energy in joule is added to a 12 g of sample of aluminum (c=0.897 J/g ◦C) to raise the temperature from 20 ◦C to 45 ◦C?

Engineering
1 answer:
MArishka [77]3 years ago
8 0

Answer:

269.1J

Explanation:

m = 12g

c = 0.897J/g°C

∆T = 45 - 20 = 25°C

H = mc∆T = 12 × 0.897 × 27 = 269.1J

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Write a complete C++ program that is made of functions main() and rShift(). The rShift() function must be a function without ret
erma4kov [3.2K]

Answer:

Explanation:

attached is an uploaded picture to support the answer.

the program is written thus;

#include<iostream>

using namespace std;

// function declaration

void rShift(int&, int&, int&, int&, int&, double&);

int main()

{

   // declare the variables

   int a1, a2, a3, a4;

   int maximum = 0;

   double average = 0;

   // inputting the numbers

   cout << "Enter all the 4 integers seperated by space -> ";

   cin >> a1 >> a2 >> a3 >> a4;

   cout << endl << "Value before shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

        << a3 << ", a4 = " << a4 << endl << endl;

   // calling rSift()

   // passing the actual parameters

   rShift(a1,a2,a3,a4,maximum,average);

   // printing the values

   cout << "Value after shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

           << a3 << ", a4 = " << a4 << endl << endl;

   cout << "Maximum value is: " << maximum << endl;

   cout << "Average is: " << average << endl;

}

// function to right shift the parameters circularly

// and calculate max, average of the numbers

// and return it to main using pass by reference

void rShift(int &n1, int &n2, int &n3, int &n4, int &max, double &avg)

{

   // calculating the max

   max = n1;

   if(n2 > max)

     max = n2;

   if(n3 > max)

     max = n3;

   if(n4 > max)

     max = n4;

   // calculating the average

   avg = (double)(n1+n2+n3+n4)/4;

   // right shifting the numbers circulary

   int temp = n2;

   n2 = n1;

   n1 = n4;

   n4 = n3;

   n3 = temp;

}

8 0
3 years ago
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postnew [5]

Answer:

<h3>Yes</h3>

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