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Nikolay [14]
3 years ago
14

Write the definition of a function printLarger, which has two int parameters and returns nothing. The function prints the larger

value of the two parameters to standard output on a single line by itself.
Engineering
1 answer:
vredina [299]3 years ago
7 0

Answer:

  1. #include <iostream>
  2. using namespace std;
  3. void printLarger(int a, int b){
  4.    
  5.    if(a > b){
  6.        cout<<a;
  7.    }else{
  8.        cout<<b;
  9.    }
  10. }
  11. int main()
  12. {
  13.    printLarger(4, 5);
  14.    return 0;
  15. }

Explanation:

The solution code is written in C++.

Firstly define a function printLarger that has two parameters, a and b with both of them are integer type (Line 5). In the function, create an if condition to check if a bigger than b, print a to terminal (Line 7-8). Otherwise print b (Line 9-10).

In the main program, test the function by passing 4 and 5 as arguments (Line 16) and we shall get 5 printed.  

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Vlad1618 [11]

Answer:

a) Internal energy

Explanation:

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Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

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FrozenT [24]

Answer:

B. Italy

Explanation:

Hope this helps :)

6 0
3 years ago
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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
zvonat [6]

Answer:

R= 1.25

Explanation:

As given the local heat transfer,

Nu_x = 0.035 Re^{0.8}_x Pr^{1/3}

But we know as well that,

Nu=\frac{hx}{k}\\h=\frac{Nuk}{x}

Replacing the values

h_x=Nu_x \frac{k}{x}\\h_x= 0.035Re^{0.8}_xPr^{1/3} \frac{k}{x}

Reynolds number is define as,

Re_x = \frac{Vx}{\upsilon}

Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity

Then replacing we have

h_x=0.035(\frac{Vx}{\upsilon})^{0.8}Pr^{1/3}kx^{-1}

h_x=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}kx^{0.8-1}

h_x=Ax^{-0.2}

<em>*Note that A is just a 'summary' of all of that constat there.</em>

<em>That is A=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}k</em>

Therefore at x=L the local convection heat transfer coefficient is

h_{x=L}=AL^{-0.2}

Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so

h=\frac{1}{L}\int\limit^L_0 h_x dx\\h=\frac{1}{L}\int\limit^L_0 AL^{-0.2}dx\\h=\frac{A}{0.8L}L^{0.8}\\h=1.25AL^{-0.2}

The ratio of the average heat transfer coefficient over the entire plate  to the local convection heat transfer coefficient is

R = \frac{h}{h_L}\\R= \frac{1.25Al^{-0.2}}{AL^{-0.2}}\\R= 1.25

3 0
3 years ago
I’m bored let’s text
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Answer:

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