PLEASE HELP!<br> I'm in the middle of a test and the teacher didn't go over the material!

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

2 years ago

A 100-ampere resistor bank is connected to a controller with conductor insulation rated 75°C. The resistors are not used in conj

Naily [24]

Answer:

answer

Explanation:

6 0

2 years ago

Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th

Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

$PV=mRT$

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

$T_1 = 38\°C$

$T_2 = 14\°C$

$\eta = 97\%$

$\dot{v} = 510m^3/kg$

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

$PV = mRT$

$\dot{m} = \frac{PV}{RT}$

$\dot{m} = \frac{0.6626*10^{5}*510}{287*311}$

$\dot{m} = 37.85kg/min$

Therfore the mass flow rate at which water condenses, then

$\eta = \frac{\dot{m_v}}{\dot{m}}$

Re-arrange to find $\dot{m_v}$

$\dot{m_v} = \eta*\dot{m}$

$\dot{m_v} = 0.97*37.85$

$\dot{m_v} = 36.72 kg/min$

PART B) Enthalpy is given by definition as,

$H= H_a +H_v$

Where,

$H_a$ = Enthalpy of dry air

$H_v$ = Enthalpy of water vapor

Replacing with our values we have that

$H=m*0.0291(38-25)+2500m_v$

$H = 37.85*0.0291(38-25)-2500*36.72$

$H = 91814.318kJ/min$

In the conversion system 1 ton is equal to 210kJ / min

$H = 91814.318kJ/min(\frac{1ton}{210kJ/min})$

$H = 437.2tons$

The cooling requeriment in tons of cooling is 437.2.

3 0

2 years ago

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

2 years ago

Write a C program that will update a bank balance. A user cannot withdraw an amount ofmoney that is more than the current balanc

GarryVolchara [31]

Answer:

Explanation:

Sample output:

BANK ACCOUT PROGRAM!

----------------------------------

Enter the old balance: 1234.50

Enter the transactions now.

Enter an F for the transaction type when you are finished.

Transaction Type (D=deposit, W=withdrawal, F=finished): D

Amount: 568.34

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 25.68

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 167.40

Transaction Type (D=deposit, W=withdrawal, F=finished): F

Your ending balance is $1609.76

Program is ending

Code to copy:

// include the necessary header files.

#include<stdio.h>

// Definition of the function

float withdraw(float account_balance, float withdraw_amount)

{

// Calculate the balace amount.

float balance_amount = account_balance - withdraw_amount;

// Check whether the withdraw amount

// is greater than 0 or not.

if (withdraw_amount > 0 && balance_amount >= 0)

{

// Assign value.

account_balance = balance_amount;

}

// return account_balance

return account_balance;

}

// Definition of the function deposit.

float deposit(float account_balance, float deposit_amount)

{

// Check whether the deposit amount is greater than zero

if (deposit_amount > 0)

{

// Update account balance.

account_balance = account_balance + deposit_amount;

}

// return account balance.

return account_balance;

}

int main()

{

// Declare the variables.

float account_balance;

float deposit_amount;

float withdrawl_amount;

char input;

// display the statement on console.

printf("BANK ACCOUT PROGRAM!\n");

printf("----------------------------------\n");

// prompt the user to enter the old balance.

printf("Enter the old balance: ");

// Input balance

scanf("%f", &account_balance);

// Display the statement on console.

printf("Enter the transactions now.\n");

printf("Enter an F for the transaction type when you are finished.\n");

// Start the do while loop

do

{

// prompt the user to enter transaction type.

printf("Transaction Type (D=deposit, W=withdrawal, F=finished): ");

// Input type.

scanf(" %c", &input);

// Check if the input is D

if (input == 'D')

{

// Prompt the user to input amount.

printf("Amount: ");

// input amount.

scanf("%f", &deposit_amount);

// Call to the function.

account_balance=deposit(account_balance,deposit_amount);

}

// Check if the input is W

if (input == 'W')

{

printf("Amount: ");

scanf("%f", &withdrawl_amount);

// Call to the function.

account_balance = withdraw(account_balance,withdrawl_amount);

}

// Check if the input is F

if (input == 'F')

{

// Dispplay the amount.

printf("Your ending balance is $%.2f\n", account_balance);

printf("Program is ending\n");

}

// End the while loop

} while(input != 'F');

return 0;

}

the picture uploaded below shows the program screenshot.

cheers, i hope this helps.

5 0

2 years ago

PLEASE HELP! I'm in the middle of a test and the teacher didn't go over the material!