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Anit [1.1K]
3 years ago
12

The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and

inside diameter of 24.0 in. If the facing operation is performed at a rotational speed = 40 rev/min, feed = 0.015 in/rev, and depth = 0.180 in, determine: (a) the cutting time to complete the facing operation, (b) the cutting speeds and metal removal rates at the beginning and end of the cut.

Engineering
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

(a) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

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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
3 years ago
A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee
Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed  = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist =  42 s

solution

we know  initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at  

so acceleration a will be

a = \frac{25-0}{8}

a = 3.125  m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 ×  t

S2 = 25  × 16

S2 = 400 m = 0.4 km  

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with  uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = \frac{S}{t}

velocity of motorist =  \frac{500}{42}  

velocity of motorist = 42.857 km/h

3 0
3 years ago
A tension test is carried out on an Al alloy specimen which has an original diameter of 0.505 in and an original gauge length of
Contact [7]

Answer:

Detailed solution is given in attached image

5 0
3 years ago
Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,
aalyn [17]

Answer:

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       System.out.println(birthMonth+"/"+birthYear);

   }

}

3 0
3 years ago
A vertical piston-cylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a c
Ann [662]

Answer:

Amount of air left in the cylinder=m_{2}=0.357 Kg

The amount of heat transfer=Q=0

Explanation:

Given

Initial pressure=P1=300 KPa

Initial volume=V1=0.2m^{3}

Initial temperature=T_{1}=20 C

Final Volume=V_{2}=0.1 m^{3}

Using gas equation

m_{1}=((P_{1}*V_{1})/(R*T_{1}))

m1==(300*0.2)/(.287*293)

m1=0.714 Kg

Similarly

m2=(P2*V2)/R*T2

m2=(300*0.1)/(0.287*293)

m2=0.357 Kg

Now calculate mass of air left,where me is the mass of air left.

me=m2-m1

me=0.715-0.357

mass of air left=me=0.357 Kg

To find heat transfer we need to apply energy balance equation.

Q=(m_{e}*h_{e})+(m_{2}*h_{2})-(m_{1}*h_{1})

Where me=m1-m2

And as the temperature remains constant,hence the enthalpy also remains constant.

h1=h2=he=h

Q=(me-(m1-m2))*h

me=m1-me

Thus heat transfer=Q=0

6 0
3 years ago
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