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shepuryov [24]
3 years ago
12

Write multiple if statements. If car_year is 1969 or earlier, print "Few safety features." If 1970 or later, print "Probably has

seat belts." If 1990 or later, print "Probably has antilock brakes." If 2000 or later, print "Probably has airbags." End each phrase with a period and a newline. Sample output for input: 1995
Engineering
1 answer:
Slav-nsk [51]3 years ago
5 0

Answer:

Explanation along with code and output results is provided below.

C++ Code:

#include <iostream>

using namespace std;

int main()

{

   int year;

   cout<<"Enter the car model year."<<endl;

   cin>>year;    

  if (year<=1969)

  {

cout<<"Few safety features."<<endl;

  }

else if (year>=1970 && year<1989)

{

cout<<"Probably has seat belts."<<endl;

}

else if (year>=1990 && year<1999)

{

cout<<"Probably has antilock brakes."<<endl;

}

else if (year>=2000)

{

cout<<"Probably has airbags."<<endl;

   }

   return 0;

}

Explanation:

The problem was to print feature messages of a car given its model year.

If else conditions are being used incorporate the logic. The code has been tested with several inputs and got correct output results.

Output:

Enter the car model year.

1961

Few safety features.

Enter the car model year.

1975

Probably has seat belts.

Enter the car model year.

1994

Probably has antilock brakes.

Enter the car model year.

2005

Probably has airbags.

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The zone where athletes may hand off baton in exchange zone is the changeover zone.

Explanation:

A relay race is a track event that involves the exchange of baton among a set of athletes. The baton is majorly exchange within the exchange zone which is 20 meters long. The exchange zone is made up on two other zones:  acceleration zone and changeover zone.

Acceleration zone is the region of the track that allows the athlete to take the next leg to start increasing speed before collecting the baton.

Changeover zone is the section where baton exchange between athletes takes place. The baton should only be exchanged within the changeover zone.

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3 years ago
What's the best way to find the load capacity of a crane? Select the best option. Call the manufacturer Ask co-workers Look at t
Orlov [11]

Answer: You need to lift a load of 15 tons (30,000 pounds) a distance of 25 feet. The distance is measured from the center pin of the crane to the center of the load. Once you determine the distance, look on that line for the largest capacity; that will indicate how many feet of boom must be extended

Explanation:

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3 years ago
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3 0
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Read 2 more answers
Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb
Nezavi [6.7K]

Answer:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- \xi_1 and \xi_2: extent of the reactions (2).

- F_{O_2}^2, F_{CH_4}^2, F_{H_2O}^2, F_{HCHO}^2 and F_{CO_2}^2: Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- F_{O_2}^2=50mol/s-\xi_1-2\xi_2: oxygen mole balance.

- F_{CH_4}^2=50mol/s-\xi_1-\xi_2: methane mole balance.

- F_{H_2O}^2=\xi_1+2\xi_2: water mole balance.

- F_{HCHO}^2=\xi_1: formaldehyde mole balance.

- F_{CO_2}^2=\xi_2: carbon dioxide mole balance.

Thus, the degrees of freedom are:

DF=7unknowns-5equations=2

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

0.900=\frac{\xi_1+\xi_2}{50mol/s}

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

0.860=\frac{F_{HCHO}^2}{50mol/s}

In such a way, one realizes that the output formaldehyde's molar flow is:

F_{HCHO}^2=0.860*50mol/s=43mol/s

Which is equal to the first reaction extent \xi_1, therefore, one computes the second one from the fractional conversion of methane as:

\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s

Now, one computes the rest of the output flows via:

- F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s

- F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s

- F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s

- F_{HCHO}^2=43mol/s

- F_{CO_2}^2=2mol/s

The total output molar flow is:

F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s

Therefore the output stream composition turns out into:

y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02

Best regards.

7 0
3 years ago
For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True
Verdich [7]

Answer:

B.

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table  for a NAND gate with 2 inputs is as follows:

0 0    1

0 1     1

1  0    1

1  1    0

The  truth table for an OR gate, is as follows:

0 0    0

0 1     1

1  0    1

1  1     1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0    1  1       1

0 1     1  0      1

1  0    0  1      1

1  1     0  0     0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

3 0
3 years ago
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