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My name is Ann [436]
2 years ago
9

What is the name of this piece?

Engineering
1 answer:
Ivanshal [37]2 years ago
7 0

Answer: block o’ metal

Explanation:

You might be interested in
A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air comp
Klio2033 [76]

Answer:

L = 46.35 m

Explanation:

GIVEN DATA

\dot m  = 0.25 kg/s

D = 40 mm

P_1 = 690 kPa

P_2 = 650 kPa

T_1 = 40° = 313 K

head loss equation

[\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m

whereh_l = \frac{ flv^2}{2D}

h_m minor loss

density is constant

v_1 = v_2

head is same so,z_1 = z_2

curvature is constant so\alpha = constant

neglecting minor losses

\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}

we know\dot m is given as= \rho VA

\rho =\frac{P_1}{RT_1}

\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3

therefore

v = \frac{\dot m}{\rho A}

V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}

V = 25.90 m/s

Re = \frac{\rho VD}{\mu}

for T = 40 Degree, \mu = 1.91*10^{-5}

Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}

Re = 4.16*10^5 > 2300 therefore turbulent flow

for Re =4.16*10^5 , f = 0.0134

Therefore

\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}

L = \frac{(P_1-P_2) 2D}{\rho f v^2}

L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}

L = 46.35 m

5 0
3 years ago
Create a package named one_dimensional_array and write a class that completes the following "OneDimensionalArrays" class. You wi
yan [13]

Answer:

The filled in codes are

1) private static int[] arr;

2)  int arr[] = new int[size_of_array];

        int increment = 100;

        for (int i = 0; i < size_of_array; i++) {

       arr[i] = increment * i;

       }

      return arr;

3) for (int i = 0; i < myArray.length; i++) {

      System.out.println(myArray[i]);

4)  OneDimensionalArrays result = new OneDimensionalArrays();

    result.createIntegers(num);

    result.printArray(arr);

Explanation:

Create a package named one_dimensional_array and write a class that completes the following "OneDimensionalArrays" class. You will complete the class by filling in code wherever you see the comment:

//******* FILL IN CODE *********

import java.util.Scanner;

public class OneDimensionalArrays {

   

   int[] createIntegers(int size_of_array)

   {

      //*******  FILL IN CODE *********

      // Your code will create an array of ints as large as specified in size_of_array

      // Fill the array in with the values: 0, 100, 200, 300, ....

      // Return the array that you just created

   }

   void printArray(int[] myArray)

   {

      //*******  FILL IN CODE *********

       // Print out your array with one number per line.  Get the size of the

       // array from the "myArray" parameter (no hard coding the size)

   }

   public static void main(String[] args) {

       Scanner keyboard = new Scanner(System.in);

       

       System.out.println("Enter size of array to create: ");

       int num = keyboard.nextInt();

       //*******  FILL IN CODE *********

       // Construct an instance of the OneDimensionalArrays class

       // Using this object instance, call createIntegers to create

       // an array of integers.  Don't forget to save the results

       // Then call the printArray method to print out the contents

       // of your array.

       }

}

Completed Code when filled in looks this way below:

import java.util.Scanner;

public class OneDimensionalArrays {

  private static int[] arr;

  int[] createIntegers(int size_of_array) {

       int arr[] = new int[size_of_array];

       int increment = 100;

       for (int i = 0; i < size_of_array; i++) {

             arr[i] = increment * i;    

       }

       return arr;

   }

   void printArray(int[] myArray) {

       for (int i = 0; i < myArray.length; i++) {

           System.out.println(myArray[i]);          

       }

   }

   public static void main(String[] args) {

       Scanner keyboard = new Scanner(System.in);

       System.out.println("Enter size of array to create: ");

       int num = keyboard.nextInt();

       OneDimensionalArrays result = new OneDimensionalArrays();

       result.createIntegers(num);

       result.printArray(arr);

   }

}

7 0
3 years ago
LOLOLOLOKOLLOLLOLOLOO STRIKER KID THINKS HES SO GOOD LLOLOLOLOLOLOLOLOLOLOOLOLOLOLOLOLOL
siniylev [52]
Rnekfkfkdkfjfnmdlwocbrig
7 0
3 years ago
Read 2 more answers
Answer?...................
torisob [31]

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.

It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

5 0
3 years ago
A strong base (caustic or alkali) is added to oils to test for: b. entrained air a)- alkalinity b)- acidity c)- contamination. d
fredd [130]

Answer:

A strong base(caustic or alkali) is added to oils test for stability

option (d) is correct

Explanation:

Formation of emulsion involves mixing together of measured volumes of specific concentrations of crude oil and sodium hydroxide (NaOH) solution. the viscosity of the emulsion and its type is determined and then this emulsion is made to rest for a long time in order to determine its stability.

7 0
3 years ago
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