Use phasor techniques to determine the impedance seen by the source given that R = 4 Ω, C = 12 μF, L = 6 mH and ω = 2000 rad/sec
1 answer:
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Answer:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
<u>A) would the 1040 steel be a possible candidate for this application</u>
<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>
stress = p / A0 = ( 18000 ) / ( ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>
Area1 = ( ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%
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.
