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krek1111 [17]
2 years ago
10

Thoughts on Anime? Whats your fav

Engineering
2 answers:
nlexa [21]2 years ago
5 0
Is miraculous ladybug one
rodikova [14]2 years ago
3 0

Answer:

My fav is jojos bizzare adventure

You might be interested in
A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 M
Readme [11.4K]

Answer:Yes,266.66 MPa

Explanation:

Given

Yield stress of material =140 MPa

Cross-section of 300\times 100 mm^2

Force(F)=8 MN

Therefore stress due to this Force(\sigma)

\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}

\sigma =266.66 \times 10^{6} Pa

\sigma =266.66 MPa

Since induced stress  is greater than Yield stress therefore Plastic deformation occurs

8 0
3 years ago
Remember from Lab 3C that Mad Libs are activities that have a person provide various words, which are then used to complete a sh
Harlamova29_29 [7]

Answer:

Python code is explained below

Explanation:

CODE(TEXT): -

9A.py:-

def sentence(): #function to read input and display

  name = input("Enter a name: ") #input name

  place = input("Enter a place: ") #input place

  number = int(input("Enter a number: ")) #input number typecasted to int

  noun = input("Enter a plural noun: ") #input plural noun

  adjective = input("Enter an adjective: ") #input adjective

  print("\n{} went to {} to buy {} different types of {}".format(name, place, number, noun)) #format is used to display o/p

  print("but unfortunately, the {} were all {} so {} went back to {} to return them.\n".format(noun, adjective,name, place))

op = "n" #initially op = n i.e. user not wanting to quit

while op != "y" : #while user does not input y

  sentence() #input words from user

  op = input("Do you want to quit [y/n]? ") #prompt to continue or quit

  if op == "y": #if yes then print goodbye and exit

      print("Goodbye")

      break

  else: #else print newline and take input

      print("")

9B.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

#then with the help of map() function we can convert all the strings into integers

numbers = list(map(int, input("Enter the input: ").split()))

sum = 0 #sum is initially 0

for i in range(len(numbers)): #loops for all the elements in the list

  sum = sum + numbers[i] #add the content of current element to sum

avg = sum/ len(numbers) #average = (sum of all elements)/ number of elements in the list

max = numbers[0] #initially max = first number

for i in range(1, len(numbers)): #loops for all remaining elements

  if numbers[i] > max : #if their content is greater than max

      max = numbers[i] #update max

print("The average and max are: {:.2f} {}".format(avg, max)) #format is used to print in formatted form

#.2f is used to print only 2 digits after decimals

9C.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

#then with the help of map() function we can convert all the strings into integers

numbers = list(map(int, input("Enter a list of numbers: ").split()))

numbers = [elem for elem in numbers if elem >= 0] #using list comprehension

#loops for all elements of the list and keep only those who are >= 0

numbers.sort() #list inbuilt function to sort items

print("The non-negative and sorted numbers are: ", end = "")

for num in numbers: #for all numbers in the list print them

  print("{} ".format(num), end = "")

7 0
3 years ago
Ordan has _ 5 8 can of green paint and _ 3 6 can of blue paint. If the cans are the same size, does Jordan have more green paint
Morgarella [4.7K]

Answer:

Jordan has more green paints

Explanation:

Given

Green = \frac{5}{8}

Blue = \frac{3}{6}

Required

Which paint does he have more?

For better understanding, it's better to convert both measurements to decimal.

For the green paint:

Green = \frac{5}{8}

Green = 0.625

For the blue paint:

Blue = \frac{3}{6}

Blue = 0.5

By comparison:

0.625 > 0.5

<em>This means that Jordan has more green paints</em>

3 0
3 years ago
Define Plastic vs elastic deformation.
Snowcat [4.5K]

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

3 0
2 years ago
Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is
ddd [48]

Answer:

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

      cout << "\n----------------------\n";

      for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

      cout << "\n----------------------\n";

      int firstDay = day-1;

      //Space print

      for(int i=1; i< firstDay; i++)

          cout << left << setw(1) << "|" << setw(2) << " ";

      int cellCnt = 0;

      // Iteration of days

      for(int i=1; i<=numDays; i++)

      {

          //Output days

          cout << left << setw(1) << "|" << setw(2) << i;

          cellCnt += 1;

          // New line

          if ((i + firstDay-1) % 7 == 0)

          {

              cout << left << setw(1) << "|";

              cout << "\n----------------------\n";

              cellCnt = 0;

          }

      }

      // Empty cell print

      if (cellCnt != 0)

      {

          // For printing spaces

          for(int i=1; i<7-cellCnt+2; i++)

              cout << left << setw(1) << "|" << setw(2) << " ";

          cout << "\n----------------------\n";

      }

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end

3 0
3 years ago
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