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VikaD [51]
3 years ago
8

6.

Engineering
1 answer:
Licemer1 [7]3 years ago
7 0

Answer:

d

none of the above

hope this is helpful

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A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0
3 years ago
By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer
Mademuasel [1]

Answer: Attached below is the well written question and solution

answer:

i) Attached below

ii) similar parameter =  \frac{V}{VoL } = 1 / Re

Explanation:

Using ;  L as characteristic length and Vo as reference velocity

i) Nondimensionalize the equations

ii) Identifying similarity parameters

the similar parameters are  = \frac{V}{VoL } = 1 / Re

Attached below is the detailed solution

7 0
3 years ago
What is Applied Science?​
andreev551 [17]

Answer:

Applied science is the application of existing scientific knowledge to practical applications, like technology or inventions. Within natural science, disciplines that are basic science develop basic information to predict and perhaps explain and understand phenomena in the natural world.

6 0
3 years ago
Please add comments to your program to explain it. Thank you!
Step2247 [10]

Answer / Explanation:

(1) We should first understand that the input filename are passed in as the first command arguments at command line, respectively.

To do this, we import the data file:

So we have,

import java.io.*;

/**

*  Makes a copy of a file.  The original file and the name of the

*  copy must be given as command-line arguments.  In addition, the

*  first command-line argument can be "-f"; if present, the program

*  will overwrite an existing file; if not, the program will report

*  an error and end if the output file already exists.  The number

*  of bytes that are copied is reported.

*/

public class CopyFile {

  public static void main(String[] args) {

     String sourceName;   // Name of the source file,  

                          //    as specified on the command line.

     String copyName;     // Name of the copy,  

                          //    as specified on the command line.

     InputStream source;  // Stream for reading from the source file.

     OutputStream copy;   // Stream for writing the copy.

     boolean force;  // This is set to true if the "-f" option

                     //    is specified on the command line.

     int byteCount;  // Number of bytes copied from the source file.

     

     /* Get file names from the command line and check for the  

        presence of the -f option.

(2)   If the command line is not one

        of the two possible legal forms, print an error message and  

        end this program. */

   

     if (args.length == 3 && args[0].equalsIgnoreCase("-f")) {

        sourceName = args[1];

        copyName = args[2];

        force = true;

     }

     else if (args.length == 2) {

        sourceName = args[0];

        copyName = args[1];

        force = false;

     }

     else {

        System.out.println(

                "Usage:  java CopyFile <source-file> <copy-name>");

        System.out.println(

                "    or  java CopyFile -f <source-file> <copy-name>");

        return;

     }

     

     /* Create the input stream.  If an error occurs, end the program. */

     

     try {

        source = new FileInputStream(sourceName);

     }

     catch (FileNotFoundException e) {

        System.out.println("Can't find file \"" + sourceName + "\".");

        return;

     }    

     /* If the output file already exists and the -f option was not

        specified, print an error message and end the program. */

   

     File file = new File(copyName);

     if (file.exists() && force == false) {

         System.out.println(

              "Output file exists.  Use the -f option to replace it.");

         return;  

    }      

     /* Create the output stream.  If an error occurs, end the program. */

     try {

        copy = new FileOutputStream(copyName);

     }

     catch (IOException e) {

        System.out.println("Can't open output file \"" + copyName + "\".");

        return;

     }

     

   (3)   /* Copy one byte at a time from the input stream to the output

        stream, ending when the read() method returns -1 (which is  

        the signal that the end of the stream has been reached).  If any  

        error occurs, print an error message.  Also print a message if  

        the file has been copied successfully.  */  

     byteCount = 0;

     

     try {

        while (true) {

           int data = source.read();

           if (data < 0)

              break;

           copy.write(data);

           byteCount++;

        }

        source.close();

        copy.close();

        System.out.println("Successfully copied " + byteCount + " bytes.");

     }

     catch (Exception e) {

        System.out.println("Error occurred while copying.  "

                                  + byteCount + " bytes copied.");

        System.out.println("Error: " + e);

     }    

  }  // end main()  

} // end class CopyFile

8 0
4 years ago
An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if t
saveliy_v [14]

Answer:

2.83 kg

Explanation:

Given:

Volume, V = 0.8 m³

gage pressure, P = 200 kPa

Absolute pressure = gage pressure + Atmospheric pressure

= 200 + 101 = 301 kPa = 301 × 10³ N/m²

Temperature, T = 23° C = 23 + 273 = 296 K

Now,

From the ideal gas equation

PV = mRT

Where,

m is the mass

R is the ideal gas constant = 287 J/Kg K. (for air)

thus,

301 × 10³ × 0.8 = m × 287 × 296

or

m = 2.83 kg

3 0
4 years ago
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