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tia_tia [17]
3 years ago
9

This function prints a proverb. The function takes a number // from the call. If that number is a 1 it prints "Now is the time /

/ for all good men to come to the aid of their party." // Otherwise, it prints "Now is the time for all good men // to come to the aid of their country." // data in: code for ending word of proverb (integer) // data out: no actual parameter altered
Engineering
1 answer:
erastova [34]3 years ago
7 0

Answer:

// This function is written in C++ Programming Language

// Comments are used for explanatory purpose

// The scope of this code segment is limited to only the required function/module

// Program starts here

#include<iostream>

using namespace std;

int main ()

{

// Main method is omitted

}

// Declare function

void proverb(int digit)

{

if(digit == 1) // check if digit is 1

{

cout << "Now is the time for all good men to come to the aid of their party.";

}

else //Otherwise

{

cout << "Now is the time for all good men to come to the aid of their country";

}

}

// End of program

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The coefficient of static friction is the ratio of the maximum static friction force (F) between the surfaces in contact before movement commences to the normal (N) force.

<h3>What is coefficient of static friction formula?</h3>
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Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume t
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Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo  =  0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

P = \frac{E}{t} = \frac{533.3 kWh}{720 h}  = 0.74 kW = 740 W

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

I =\frac{P}{V} =\frac{740W}{120V} = 6.2A

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = \frac{533.3-490}{533.3} = 0.081*100= 8.1%

⇒ ΔE(%) = 8.1%

%(E) =\frac{533.3-490}{533.3}  = 0.081 * 100 = 8.1%

4 0
4 years ago
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