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kykrilka
25 days 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:
alex41 [274]25 days ago
5 0

Answer:

The following includes the explanation, code, and resulting outputs.

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 challenge involved displaying feature messages for a car based on its model year.

Logical conditions were integrated into the coding. The implementation has been verified with multiple inputs producing the expected 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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Given num_rows and num_cols, print a list of all seats in a theater. Rows are numbered, columns lettered, as in 1A or 3E. Print
Viktor [230]

Answer:

This is the solution code in Python:

  1. alphabets = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J']
  2. user_input = input("Enter number of rows and columns: ")
  3. myArr = user_input.split(" ")
  4. num_rows = int(myArr[0])
  5. num_cols = int(myArr[1])
  6. seats = []
  7. for i in range(num_rows):
  8. row = []
  9. for j in range(num_cols):
  10. row.append(alphabets[j])
  11. seats.append(row)
  12. output = ""
  13. for i in range(len(seats)):
  14. for j in range(len(seats[i])):
  15. output += str(i + 1) + seats[i][j] + " "
  16. print(output)

Explanation:

Initially, we create a small list of alphabets from A to J (Line 1).

We then request the user to enter the number of rows and columns (Line 3). Given that the input comes as a string (e.g., "2 3"), we utilize the split() method to separate the numbers into individual items in a list (Line 4). The first item (row number) is assigned to variable num_rows, while the second item (column number) goes to num_cols.

Subsequently, we construct the seats list with a nested for-loop (Lines 10-15). Once the seats list is formed, another nested for-loop generates the required output string as per the question (Lines 19-21).

Finally, the output is printed (Line 23). For example, an input of 2 3 results in the output:

1A 1B 1C 2A 2B 2C

8 0
1 month ago
Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have fou
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4 0
27 days ago
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
pantera1 [220]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

The half-life for the specified RC circuit can be expressed as

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

The circuit has a resistance of 40 ohms, and by adding a new resistor of 48 ohms, the total resistance becomes 40 + 48 = 88 ohms.

Thus, the new half-life is

t'_{1\2} =R'Cln2

Now, divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

After substituting all values, we can calculate the revised half-life

t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
25 days ago
The rigid beam is supported by a pin at C and an A992 steel guy wire AB of length 6 ft. If the wire has a diameter of 0.2 in., d
Mrrafil [253]

Answer:

Change in length = 0.0913 in

Explanation:

Given data:

Length = 6 ft

Diameter = 0.2 in

Load w = 200 lb/ft

Solution:

We start by applying the equilibrium moment about point C, expressed as

∑M(c) = 0.............1

This can be used to find the force in AB.

10× 200 × ( 5) - (T cos(30)) × 10 = 0

Solving gives us

Tension in wire T(AB) = 1154.7 lb

We also know the modulus of elasticity for A992 is

E = 29000 ksi

And the area will be

Area = \frac{\pi }{4}\times 0.2^2

The change in length is expressed as

Change in length = \frac{PL}{AE}.........2

Substituting values results in

Change in length = \frac{1154.7 \times 6 \times 12}{\frac{\pi }{4}\times 0.2^2 \times 29000 \times 1000}

Change in length = 0.0913 in

8 0
26 days ago
An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,
Mrrafil [253]

Answer:

Volume change percentage is 2.60%

Water level increase is 4.138 mm

Explanation:

Provided data

Water volume V = 500 L

Initial temperature T1 = 20°C

Final temperature T2 = 80°C

Diameter of the vat = 2 m

Objective

We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

Thus,

E = -\frac{dp}{dV/V}................1

And -\frac{dV}{V} = \frac{d\rho}{\rho}............2

Here, ρ denotes density. The density at 20°C = 998 kg/m³.

The density at 80°C = 972 kg/m³.

Plugging in these values into equation 2 gives

-\frac{dV}{V} = \frac{d\rho}{\rho}

-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}

dV = 0.0130 m³

Therefore, the percentage change in volume will be

dV % = -\frac{dV}{V}  × 100

dV % = -\frac{0.0130}{500*10^{-3} }  × 100

dV % = 2.60 %

Hence, the percentage change in volume is 2.60%

Initial volume v1 = \frac{\pi }{4} *d^2*l(i)................3

Final volume v2 = \frac{\pi }{4} *d^2*l(f)................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 =  \frac{\pi }{4} *d^2*(l(f)-l(i))

dV = \frac{\pi }{4} *d^2*dl

Substituting all values yields

0.0130 = \frac{\pi }{4} *2^2*dl

Thus, dl = 0.004138 m.

Consequently, the water level rises by 4.138 mm.

8 0
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