Implement a program that manages shapes. Implement a class named Shape with a method area() which returns the double value 0.0. Implement three derived classes named Rectangle, Square, and Circle. Declare necessary properties in each including getter and setter function and a constructor that sets the values of these properties. Override the area() function in each by calculating the area using the defined properties in that class.Using Java to write a program that repeatedly shows the user a menu to create one of the three main shapes or to print the shapes created so far. If the user selects to create a new shape, the program prompts the user to enter the values for the size of the selected shape. The shape is then stored in an array. If the user selects to print the current shapes, print the name and the total area of each shape to the console.Hint: You may limit the size of the array to 10.

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, [tex]$h_0=2.2$[/tex] inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

[tex]$\Delta h = H_0-h_f=\mu^2R$[/tex]

     [tex]$=0.2^2 \times 14 = 0.56$[/tex] inches

[tex]$h_f = 2.2 - 0.56$[/tex]

     = 1.64 inches

Roll strip contact length (L) = [tex]$\sqrt{R(h_0-h_f)}$[/tex]

                                             [tex]$=\sqrt{14 \times 0.56}$[/tex]

                                             = 2.8 inches

Absolute value of true strain, [tex]$\epsilon_T$[/tex]

[tex]$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$[/tex]

Average true stress, [tex]$\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$[/tex] Psi

Roll force, [tex]$L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$[/tex]

                                 = 788,900 lb

For SI units,

Power = [tex]$\frac{2 \pi FLN}{60}$[/tex]  

           [tex]$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$[/tex]

           = 10399.81168 W

Horse power = 13.9357

C. Semiconductors.

They are made up of what is called a porous silicon material that is very light and extremely heat resistant.

Answer:

composites

Explanation:

Answer:

$Monitor

Explanation:

The command that would be used when running a test bench to monitor variables or signals ( i.e. changes in the values of specific variables and signa)

is the $Monitor command

This command is also used to monitor the varying values of signals during simulation.

Answer:

A

Explanation:

2/10 = 1/5 = 0.2

Answer:

ICC

Explanation:

The International Building Code (IBC) is a model building code developed by the International Code Council (ICC). It has been adopted for use as a base code standard by most jurisdictions in the United States.

Answer:

organizational goals

Explanation:

Answer:

Explanation:

Pie charts generally should have no more than eight segments.

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

h = 389.55213360 W/m²

h ≅ 389.6 W/m²

Answer:

er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)

Explanation:

er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)er of passes through zero

Example:

Input data:

10

1 3 -2 -6 4 10 1 -5 4 1 Output data:

4

Explanation:

There are 4 sign changes between two consecutive samples:

3 -2 -> sign change (first zero crossing)

-6 4 -> sign change (second zero crossing)

Answer:

Technical communication is a means to convey scientific, engineering, or other technical information.[1] Individuals in a variety of contexts and with varied professional credentials engage in technical communication. Some individuals are designated as technical communicators or technical writers. These individuals use a set of methods to research, document, and present technical processes or products. Technical communicators may put the information they capture into paper documents, web pages, computer-based training, digitally stored text, audio, video, and other media. The Society for Technical Communication defines the field as any form of communication that focuses on technical or specialized topics, communicates specifically by using technology or provides instructions on how to do something.[2][3] More succinctly, the Institute of Scientific and Technical Communicators defines technical communication as factual communication, usually about products and services.[4] The European Association for Technical Communication briefly defines technical communication as "the process of defining, creating and delivering information products for the safe, efficient and effective use of products (technical systems, software, services)".[5]

Whatever the definition of technical communication, the overarching goal of the practice is to create easily accessible information for a specific audience.[6]

Answer:

a. True

Explanation:

The study of fluids in a state of rest or in motion and the forces involved in it is called fluid mechanics. Fluid mechanics has a wide range of applications in the field of mechanical engineering as well as civil engineering.

When we study the flow of fluid between any two flat plates that is indefinitely flat and is parallel, the flow of the fluid is known as plane Poiseulle flow. The profile of a plane Poiseulle flow is parabolic.

The velocity profile of a plane Poiseulle flow is :

[tex]$\frac{u(y)}{U_{max}}=1-\left(\frac{2y}{h}\right)^2$[/tex]

Thus the answer is TRUE.

Answer:

M/A = 0.425 g/cm²

Explanation:

Given the following data;

Mass = 192 grams

Dimensions = 7 * 10 inches.

To find the mass per unit area;

First of all, we would determine the area of the lead sheet;

Area = 7 * 10

Area = 70 in²

Conversion:

1 square inch = 6.452 square centimeters

70  inches = 70 * 6.452 = 451.64 square centimeters

Next, we find the mass per unit area;

M/A = 192/451.64

M/A = 0.425 g/cm²

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

                                 = 500 - 2(20)

                                 = 460 mm

So, internal radius, r = 230 mm

                                 = 0.23 m

Density of molten metal, ρ = [tex]$7.2 \ g/cm^3$[/tex]

                                                  = [tex]$7200 \ kg/m^3$[/tex]

The height of pouring cavity above parting surface is h = 300 mm

                                                                                                  = 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

[tex]$F=\rho g A h$[/tex]

Area, A [tex]$=2 \pi r^2$[/tex]

            [tex]$=2 \pi (0.23)^2$[/tex]

            [tex]$=0.3324 \ m^2$[/tex]

[tex]$F=\rho g A h$[/tex]

   [tex]$=7200 \times 9.81 \times 0.3324 \times 0.3$[/tex]

     = 7043.42 N

Answer:

a) | v2 | ,  β2   ( load, bus voltage at bus 2 )

  p1 ,  q1 ( slack, bus power at bus 1 )

b) q2 , β2  

  p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

a) Identify the variables in the solution vector assume Bus 2 is a load bus

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 |  and β1 been specified while p1 and q1 are not specified

Hence the variables in the solution

= | v2 | ,  β2   ( load, bus voltage at bus 2 )

  p1 ,  q1 ( slack, bus power at bus 1 )

b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 |  and β1

unspecified : p1 and q1

Hence the variables in the solution

= q2 , β2  

  p1 and q1 ( slack, bus power at bus 1 )

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

[tex]E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%[/tex]

Answer:

W18 * 106

Explanation:

Section modulus of wide flange = 200 m^3

Determine the value of the lightest beam possible

The lightest beam possible that will satisfy the given condition will  have a section modulus ≥ 200m^3 ( note: it will also be the nearest to 200 in^3 )

From Beam Table ; The Lightest beam with its section modulus( 204 in^3) > 200in^3  is   W18 * 106

Answer:

- the undrained friction angles for the soil is 25.02°

- the drained friction angles for the soil is 18.3°

Explanation:

Given the data in the question;

First we determine the major principle stress using the express;

σ₁ = σ₃ + (Δσ[tex]_d[/tex] )[tex]_f[/tex]

where σ₃ is the total minor principle stress at failure ( 15 lb/in² )

(Δσ[tex]_d[/tex] )[tex]_f[/tex] is the deviator stress ( -9 lb/in² )

so

σ₁ = 15 lb/in² + 22 lb/in²

σ₁ = 37 lb/in²

Now, we calculate the consolidated-undrained friction angle as follows;

∅ = sin⁻¹[ (σ₁ - σ₃ ) / ( σ₁ + σ₃ ) ]

∅ = sin⁻¹[ (37 - 15 ) / ( 37 + 15 ) ]

∅ = sin⁻¹[ 22 / 52  ]

∅ = sin⁻¹[ 0.423 ]

∅ = 25.02°

Therefore, the undrained friction angles for the soil is 25.02°

-  The drained friction angles for the soil;

∅ = sin⁻¹[ (σ₁ - σ₃ ) / ( σ₁ + σ₃ - 2(Δσ[tex]_d[/tex] )[tex]_f[/tex] ) ]

so we substitute

∅ = sin⁻¹[ (37 - 15 ) / ( 37 + 15 - 2( -9 ) ]

∅ = sin⁻¹[ 22 / ( 37 + 15 + 18 ) ]  

∅ = sin⁻¹[ 22 / 70 ]

∅ = sin⁻¹[ 0.314 ]

∅ = 18.3°

Therefore, drained friction angles for the soil is 18.3°