A continuous function y = ƒ(x) is known to be negative at x = 0 and positive at x = 1. Why does the equation ƒ(x) = 0 have at le
1 answer:
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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.
Answer:
The correct option is;
c. the exergy of the tank can be anything between zero to P₀·V
Explanation:
The given parameters are;
The volume of the tank = V
The pressure in the tank = 0 Pascal
The pressure of the surrounding = P₀
The temperature of the surrounding = T₀
Exergy is a measure of the amount of a given energy which a system posses that is extractable to provide useful work. It is possible work that brings about equilibrium. It is the potential the system has to bring about change
The exergy balance equation is given as follows;
Where;
X₂ - X₁ is the difference between the two exergies
Therefore, the exergy of the system with regards to the environment is the work received from the environment which at is equal to done on the system by the surrounding which by equilibrium for an empty tank with 0 pressure is equal to the product of the pressure of the surrounding and the volume of the empty tank or P₀ × V less the work, exergy destroyed, while taking into consideration the change in heat of the system
Therefore, the exergy of the tank can be anything between zero to P₀·V.
The composition of gas in the feed, the percentage conversion and the
theoretical yield are combined to give the product stream composition.
Response:
The composition of gas in the product stream are;
- HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h
The amount of oxygen used = 30% exceeding the theoretical amount
Number of moles of hydrochloric acid = 4 kmol/h
Percentage conversion = 80%
Required:
The composition of the gas in the product feed.
Solution;
The given reaction is; 4HCl + O₂ 2Cl₂ + 2H₂O
Which gives;
Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h
Which gives;
Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h
Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Similarly;
Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h
The composition of the production stream is therefore;
- <u>HCl: 0.4 kmol/h</u>
- <u>Cl₂: 1.6 kmol/h</u>
- <u>H₂O: 1.6 kmol/h</u>
- <u>O₂: 0.5 kmol/h</u>
Learn more about theoretical and actual yield here: