. A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The allowable shear stress for the material of the shaft is 42 MPa.

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.

Answer:

r=0.31

Ф=18.03°

Explanation:

Provided:

Original diameter of bar = 75 mm

Diameter post-cutting = 73 mm

Average diameter of the bar d= (75+73)/2=74 mm

Average length of uncut chip = πd

Average length of uncut chip = π x 74 =232.45 mm

Thus, cutting ratio r

  r=0.31

Therefore, the cutting ratio equals 0.31.

Now, the shearing angle is given as

Next by substituting the values

\

Ф=18.03°

Concluding, the shearing angle is 18.03°.

Respuesta:

La temperatura máxima que se puede medir (en °C) es 14170.27°C

Explicación:

Los RTDs son termómetros compuestos de metales cuya resistencia aumenta con la temperatura.

Para un RTD de Clase A, l, Alpha = 0.00385.

La fórmula para el RTD es

Rt = Ro ( 1 + alpha x t)

Donde

Rt es la resistencia a la temperatura t°C,

Ro es la resistencia a 0°C

Alpha es un coeficiente de temperatura constante para un RTD de clase A.

Aquí, Rt = 1000ohms,

Ro se considera como Ra = 18Ohms

Por lo tanto,

1000 = 18 ( 1 + 0.00385t)

Dividiendo ambos lados por 18

1 + 0.00385t = 1000/18

0.00385t = 55.55 - 1

0.00385t = 54.55

t = 54.55/0.00385

t = 14170.27°C

It is not feasible to install the wires within the conduit. Explanation: The given dimensions show that the total area is 2.04 square inches while the wires occupy 0.93 square inches. The maximum allowable fill for the conduit is 40%. To determine if placement is possible, compute the conduit’s area at 40% which equates to 0.816 square inches, less than the required area of the wires.

Response:

a) 4 kg/s

b) 99.61 °C

Rationale:

Refer to the pictures provided.