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Kitty [74]
3 years ago
5

Which of the following is a valid reason why a scientist might reject a scientific theory

Engineering
2 answers:
Flauer [41]3 years ago
6 0
C I think hope it helps :)
Gnesinka [82]3 years ago
4 0

Answer:

C. New evidence Contradicts it

Explanatine

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Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is be
hammer [34]

Answer:

A) i) 984.32 sec

ii) 272.497° C

B) It has an advantage

C) attached below

Explanation:

Given data :

P = 2700 Kg/m^3

c = 950 J/kg*k

k = 240 W/m*K

Temp at which gas enters the storage unit  = 300° C

Ti ( initial temp of sphere ) = 25°C

convection heat transfer coefficient ( h ) = 75 W/m^2*k

<u>A) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

First step determine the Biot Number

characteristic length( Lc ) = ro / 3 = 0.0375 / 3 = 0.0125

Biot number ( Bi ) = hLc / k = (75)*(0.0125) / 40 = 3.906*10^-3

Given that the value of the Biot number is less than 0.01 we will apply the lumped capacitance method

attached below is a detailed solution of the given problem

<u>B) The physical properties are copper</u>

Pcu = 8900kg/m^3)

Cp.cu = 380 J/kg.k

It has an advantage over Aluminum

C<u>) Determine how long it takes a sphere near the inlet of the system to accumulate 90% of the maximum possible energy and the corresponding temperature at the center of sphere</u>

Given that:

P = 2200 Kg/m^3

c = 840 J/kg*k

k = 1.4 W/m*K

3 0
3 years ago
Switches are placed only in the _ of a circuit?
postnew [5]

the switch should always be placed immediately adjacent to the non-grounded terminal of the power supply.

4 0
3 years ago
Opposition to current flow, restricts or resists current flow
BlackZzzverrR [31]

Answer:

The answer is c-resistance

8 0
3 years ago
The screw of shaft straightener exerts a load of 30 as shown in Figure . The screw is square threaded of outside diameter 75 mm
kykrilka [37]

Answer:

See calculation below

Explanation:

Given:

W = 30 kN = 30x10³ N

d = 75 mm

p = 6 mm

D = 300 mm

μ = tan Φ = 0.2

<u><em>1. Force required at the rim of handwheel </em></u>

Let P₁ = Force required at the rim of handwheel

Inner diameter or core diameter of the screw = dc = do - p = 75 - 6 = 69 mm

Mean diameter of screw:    *d = \frac{do + dc}{2} = (75 + 69) / 2 = 72 mm

and

tan α = p / πd  =  6 / (π x 72)  =  0.0265

∴ Torque required to overcome friction at he threads is  T = P x d/2

T = W tan (α + Ф) d/2

T =  W(\frac{tan \alpha + tan \theta}{1 - tan \alpha + tan \theta } ) * \frac{d}{2}

T = 30x10³ * ((0.0265 + 0.2) / (1 - 0.0265 x 0.2)) x 72/2

T = 245,400 N-mm

We know that the torque required at the rim of handwheel (T)

245,400 = P1 x D/2 = P1 x (300/2) = 150 P1

P1 = 245,400 / 150

P1 = 1636 N

<u><em>2. Maximum compressive stress in the screw</em></u>

                         30x10³

Qc = W / Ac = -------------- = 8.02 N/mm²

                      π/4 * 69²

Qc = 8.02 MPa

Bearing pressure on the threads (we know that number of threads in contact with the nut)

n = height of nut / pitch of threads = 150 / 6 = 25 threads

thickness of threads, t = p/2 = 6/2 = 3 mm

bearing pressure on the threads = Pb = W / (π d t n)

Pb = 30 x 10³ / (π * 72 * 3 * 25)

Pb = 1.77 N/mm²

Max shear stress on the threads = τ = 16 T / (π dc³)

τ = (16 * 245,400) / ( π * 69³ )

τ = 3.8 M/mm²

*the mean dia of the screw (d) = d = do - p/2 = 75 - 6/2 = 72

∴max shear stress in the threads τmax = 1/2 * sqrt(8.02² + (4 * 3.8²))

τmax = 5.5 Mpa

<u><em>3. efficiency of the straightener</em></u>

<u><em></em></u>

To = W tan α x d/2 = 30x10³ * 0.0265 * 72/2 = 28,620 N-mm

∴Efficiency of the straightener is η =  To / T = 28,620 / 245,400

η = 0.116 or 11.6%

4 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
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