Please help ASAP!!

Exercise 6.4.8: Sum Two Number

kaheart [24]

Below is the program with function that takes two arguments and returns their sum.

Coding part:

def add_num(a1,b1): /*function for addition*/

sum=a1+b1

return sum /*return value*/

num1z1=33 /*variable declaration*/

num2z2=78

print("The sum is",add_num(num1z1,num2z2)) /*call the function*/

What is Function?

A function is a piece of code that performs a specific task. It is callable and reusable multiple times. You can pass data to a function, and it can return data to you. Many programming languages include built-in functions that can be found in their libraries, but you can also write your own.

To learn more about Function, visit: brainly.com/question/17216645

#SPJ1

3 0

1 year ago

Write a function which multiplies the values in odd position values by 10. Odd positions in this case refers to the first value

xxMikexx [17]

Answer:

Using linkedlist on C++, we have the program below.

Explanation:

#include<iostream>

#include<cstdlib>

using namespace std;

//structure of linked list

struct linkedList

{

int data;

struct linkedList *next;

};

//print linked list

void printList(struct linkedList *head)

{

linkedList *t=head;

while(t!=NULL)

{

cout<<t->data;

if(t->next!=NULL)

cout<<" -> ";

t=t->next;

}

//insert newnode at head of linked List

struct linkedList* insert(struct linkedList *head,int data)

{

linkedList *newnode=new linkedList;

newnode->data=data;

newnode->next=NULL;

if(head==NULL)

head=newnode;

else

{

struct linkedList *temp=head;

while(temp->next!=NULL)

temp=temp->next;

temp->next=newnode;

}

return head;

}

void multiplyOddPosition(struct linkedList *head)

{

struct linkedList *temp=head;

while(temp!=NULL)

{

temp->data = temp->data*10; //multiply values at odd position by 10

temp = temp->next;

//skip odd position values

if(temp!= NULL)

temp = temp->next;

}

int main()

{

int n,data;

linkedList *head=NULL;

// create linked list

head=insert(head,20);

head=insert(head,5);

head=insert(head,11);

head=insert(head,17);

head=insert(head,23);

head=insert(head,12);

head=insert(head,4);

head=insert(head,21);

cout<<"\nLinked List : ";

printList(head); //print list

multiplyOddPosition(head);

cout<<"\nLinked List After Multiply by 10 at odd position : ";

printList(head); //print list

return 0;

}

5 0

3 years ago

A 304 stainless steel (yield strength 30 ksi) cylinder has an inner diameter of 4 in and a wall thickness of 0.1 in. If it is su

Natasha_Volkova [10]

Answer:

The value we got is 1.423 ksi which is less than 30 and so the material hasn't failed and yielding hasn't occured.

Explanation:

This is a cylindrical thin walled vessel. Thus;

Hoop stress; σ1 = pr/t

Where ;

p is internal pressure

r_o is radius = 4/2 = 2 in

t is thickness of wall = 0.1 in

Thus;

σ1 = 70 x 2/0.1 = 1400 ksi

Longitudinal stress; σ2 = pr/2t

σ2 = 70 x 2/(0.1 x 2) = 700 ksi

Now, we want to find the normal stress. The inner radius of the circle will be; r_i = r_o - t = 2 - 0.1 = 1.9 in

So, normal stress by axial force is given by;

σ_fx = F/A = F/(π(r_o² - r_i²))

We are given that F = 500 lb

σ_fx = 500/(π(2² - 1.9²))

σ_fx = 408.1 ksi

We can now find the torsion from the formula;

τ = (T•r_o)/J

We are given that T = 70 lb.ft = 70 x 12 lb.in = 840 lb.in

J is the polar moment of inertia and has a formula; ((π/2)(r_o⁴ - r_i⁴))

So,J = ((π/2)(2⁴ - 1.9⁴)) = 4.662 in⁴

Thus,τ_xy = (840 x 2)/4.662 = 360.4 ksi

σ1 is in the y direction and σ2 is in the x direction. Thus;

σ_x = σ1 + σ_fx = 700 + 408.1 = 1108.1 ksi

Also, σ_y = σ1 = 1400 ksi

Now distortion energy can be expressed as;

σ_Y² = σ_x - σ_x•σ_y + (σ_y)² + 3(τ_xy)²

Plugging in the relevant values, we obtain ;

σ_Y² = 1108.1² - (1108.1*1400) + 1400² + 3(360.4)²

So, σ_Y² = 2.02 x 10^(6) psi²

σ_Y = √2.02 x 10^(6)

σ_Y = 1423 psi = 1.423 ksi

The question says the yield strength of the material is 30 ksi.

The value we got is less than 30 and so the material hasn't failed and yielding hasn't occured.

8 0

3 years ago

If aligned and continuous carbon fibers with a diameter of 9.90 micron are embedded within an epoxy, such that the bond strength

Zielflug [23.3K]

I have no idea what it is

8 0

3 years ago

forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fue

Zina [86]

Answer: 131.75minutes

Explanation:

First if all, we've to find the density of liquid which will be:

= Specific gravity × Density to pure water

= 0.91 × 8.34lb/gallon

= 7.59lb/gallon

Then, the volume that's required to fill the tank will be:

= Load limit/Density of fluid

= 40000/7.59

= 5270.1gallon

Now, the time taken will be:

= V/F

= 5270.1/40

= 131.75min

It'll take 131.75 minutes to fill the tank in the truck.

5 0

3 years ago