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VMariaS [17]
3 years ago
13

Best use for a suspension bridge

Engineering
1 answer:
babunello [35]3 years ago
4 0

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

You might be interested in
Which of the following allows team members to visualize a design model from a variety of perspectives?
julsineya [31]

Answer: from what i know im pretty sure its isometrics or sketches im certain its sketches but not 100%

Explanation: A sketch is a rapidly executed freehand drawing that is not usually intended as a finished work. A sketch may serve a number of purposes: it might record something that the artist sees, it might record

8 0
3 years ago
Read 2 more answers
Which allows a user to run applications on a computing device? Group of answer choices Application software CSS Operating system
sveticcg [70]

Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user.  The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

Cheers.

4 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 air-standard dual cycle has a compression ratio of 9.1 and displacement of Vd = 2.2 L. At the beginning of compression, p1 =
jok3333 [9.3K]

Answer:

a) T₂ is 701.479 K

T₃ is 1226.05 K

T₄ is 2350.34 K

T₅ is 1260.56 K

b) The net work of the cycle in kJ is 2.28 kJ

c) The power developed is 114.2 kW

d) The thermal efficiency, \eta _{dual} is 53.78%

e) The mean effective pressure is 1038.25 kPa

Explanation:

a) Here we have;

\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}}  \right )^{\gamma -1} = \left (r  \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}}  \right )^{\frac{\gamma -1}{\gamma }}

Where:

p₁ = Initial pressure = 95 kPa

p₂ = Final pressure =

T₁ = Initial temperature = 290 K

T₂ = Final temperature

v₁ = Initial volume

v₂ = Final volume

v_d = Displacement volume =

γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air

r = Compression ratio = 9.1

Total heat added = 4.25 kJ

1/4 × Total heat added = c_v \times (T_3 - T_2)

3/4 × Total heat added = c_p \times (T_4 - T_3)

c_v = Specific heat at constant volume = 0.718×2.821× 10⁻³

c_p = Specific heat at constant pressure = 1.005×2.821× 10⁻³

v₁ - v₂ = 2.2 L

\left \frac{v_{1}}{v_{2}}  \right =r  \right = 9.1

v₁ = v₂·9.1

∴ 9.1·v₂ - v₂ = 2.2 L  = 2.2 × 10⁻³ m³

8.1·v₂ = 2.2 × 10⁻³ m³

v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³

v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³

Plugging in the values, we have;

{T_{2}}= T_{1} \times \left (r  \right )^{\gamma -1}  = 290 \times 9.1^{1.4 - 1} = 701.479 \, K

From;

\left (\frac{p_{2}}{p_{1}}  \right )^{\frac{\gamma -1}{\gamma }}= \left (r  \right )^{\gamma -1} we have;

p_{2} = p_{1}} \times \left (r  \right )^{\gamma } = 95 \times \left (9.1  \right )^{1.4} = 2091.13 \ kPa

1/4×4.25 =  0.718 \times 2.821 \times  10^{-3}\times (T_3 - 701.479)

∴ T₃ = 1226.05 K

Also;

3/4 × Total heat added = c_p \times (T_4 - T_3) gives;

3/4 × 4.25 = 1.005 \times 2.821 \times  10^{-3} \times (T_4 - 1226.05) gives;

T₄ = 2350.34 K

\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}}  \right )^{\gamma -1} = \left (\frac{r}{\rho }  \right )^{\gamma -1}

\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92

T_{5} =  \frac{T_{4}}{\left (\frac{r}{\rho }  \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 }  \right )^{1.4-1}} =1260.56 \ K

b) Heat rejected =  c_v \times (T_5 - T_1)

Therefore \ heat \ rejected =  0.718 \times 2.821 \times  10^{-3}\times (1260.56 - 290) = 1.966 kJ

The net work done = Heat added - Heat rejected

∴ The net work done = 4.25 - 1.966 = 2.28 kJ

The net work of the cycle in kJ = 2.28 kJ

c) Power = Work done per each cycle × Number of cycles completed each second

Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second

Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW

d)

Thermal \ efficiency, \, \eta _{dual} =  \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%

The thermal efficiency, \eta _{dual} = 53.78%

e) The mean effective pressure, p_m, is found as follows;

p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa

The mean effective pressure = 1038.25 kPa.

3 0
4 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Sedaia [141]

Answer:

a) 254.6 GPa

b) 140.86 GPa

Explanation:

a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;

Ec(u) = EmVm + EpVp

To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,

Vm + Vp = 1

Vm = 1 - 0.63

Vm = 0.37

In the first equation,

Where

Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,

The modulus of elasticity upper-bound is,

Ec(u) = EmVm + EpVp

Ec(u) = (68 x 0.37) + (380 x 0.63)

Ec(u) = 254.6 GPa.

b) Considering the express of rule of mixtures for lower bound;

Ec(l) = (EmEp)/(VmEp + VpEm)

Substituting values into the equation,

Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

Ec(l) = 25840/183.44

Ec(l) = 140.86 GPa

6 0
4 years ago
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