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pickupchik [31]
3 years ago
15

Two parts are to be assembled in a way that if one part fails, the entire assembly fails. Each of the parts have undergone exten

sive individual analysis and you are provided with statistics on the stresses that are expected. Specifically, you are provided with the mean and standard deviation of the stress, and the yield stress of the material for each part. Failure of a part will be defined as a part experiencing stress exceeding the yield stress. What is the probability that the assembly fails? (Hint: Use Table A-10 in the textbook to find the probability of failure for each part.)
Part #1: σnom = 260 MP a, U (σ) = 25 MP a, σyield = 310 MPa

Part #2: σnom = 475 MP a, U (σ) = 5 MP a, σyield = 490 MPa

Engineering
1 answer:
e-lub [12.9K]3 years ago
5 0

Answer:

attached below

Explanation:

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Fibonacci sequence has many applications in Computer Science. Write a program to generate Fibonacci numbers as many as desired.
VikaD [51]

Answer:

The Python Code for Fibonacci Sequence is :

# Function for nth Fibonacci number  

def Fibonacci(n):  

if n<0:  

 print("Incorrect input")  

# First Fibonacci number is 0  

elif n==0:  

 return 0

# Second Fibonacci number is 1  

elif n==1:  

 return 1

else:  

 return Fibonacci(n-1)+Fibonacci(n-2)  

# Driver Program  

print(Fibonacci(9))  

Explanation:

The Fibonacci numbers are the numbers in the following integer sequence.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

Fn = Fn-1 + Fn-2

with seed values

F0 = 0 and F1 = 1.

8 0
3 years ago
Read 2 more answers
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

Explanation: Given that the relation between pressure and volume is

PV^n = constant.

That is, P1V1^n = P2V2^n

P1 = P2 × ( V2/V1 )^n

If the initial volume V1 = 0.1 m3,

the final volume V2 = 0.04 m3, and

the final pressure P2 = 2 bar. 

A.) When n = 0

Substitute all the parameters into the formula

(V2/V1)^0 = 1

Therefore, P2 = P1 = 2 bar

Work = ∫ PdV = constant × dV

Work = 2 × 10^5 × [ 0.04 - 0.1 ]

Work = 200000 × - 0.06

Work = - 12000J

Work = - 12 kJ

B.) When n = 1

P1 = 2 × (0.04/0.1)^1

P1 = 2 × 0.4 = 0.8 bar

Work = ∫ PdV = constant × ∫dV/V

Work = P1V1 × ln ( V2/V1 )

Work = 0.8 ×10^5 × 0.1 × ln 0.4

Work = - 7330.3J

Work = -7.33 kJ

C.) When n = 1.3

P1 = 2 × (0.04/0.1)^1.3

P1 = 0.6077 bar

Work = ∫ PdV

Work = (P2V2 - P1V1)/ ( 1 - 1.3 )

Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

Work = (8000 - 6080)/ -0.3

Work = -1920/0.3

Work = -6400 J

Work = -6.4 kJ

5 0
4 years ago
C++ - Green Crud Fibonacci programThe following program is to be written with a loop. You are to write this program three times
Fynjy0 [20]

Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

    //Initial crud size

    int init = 0;

    int newCrud;

    int next=0;

    //Number of days to simulate

    int no_days;

    int day;

    cout << "Enter initial amount of green crud: ";

    cin >> newCrud;

    cout << "Enter number of days to simulate: ";

    cin >> no_days;

    for (day = 10; day<=no_days; day++)

    {

         if (day % 10 == 0)

         {

             next = newCrud + init;

         }

             newCrud = init;

             init = next;

    }

    if (no_days < 5)

    cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

    << newCrud << " pounds.";

    else

    cout << "On day " << no_days << " you have " << init

    << " pounds of green crud." << endl;

    cout << "\nWould you like to continue? (y or n): ";

    cin >> ans;

         return 0;

}

Output:

         Enter initial amount of green crud: 5

         Enter number of days to simulate: 220

    On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans='y';

    while (ans == 'Y' || ans == 'y')

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud:

         ";

         cin >> newCrud;

         cout << "Enter number of days to simulate:

         ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         cout << "On day " << no_days << " you have "

         << init

         << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or

         n): ";

         cin >> ans;

    }

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans;

    do

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud: ";

         cin >> newCrud;

         cout << "Enter number of days to simulate: ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         

         cout << "On day " << no_days << " you have " <<

         init << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or n):

         ";

         cin >> ans;

    } while (ans == 'Y' || ans == 'y');

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

7 0
3 years ago
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