Answer:

**Answer:**

Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.

**Explanation:**

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Whats are the best choice of cooling towers for electeic generating power plant and why?

IN JAVA,Knapsack ProblemThe file KnapsackData1.txt and KnapsackData2.txt are sample input filesfor the following Knapsack Problem that you will solve.KnapsackData1.txt contains a list of four prospective projects for the upcoming year for a particularcompany:Project0 6 30Project1 3 14Project2 4 16Project3 2 9Each line in the file provides three pieces of information:1) String: The name of the project;2) Integer: The amount of employee labor that will be demanded by the project, measured in work weeks;3) Integer: The net profit that the company can expect from engaging in the project, measured in thousandsof dollars.Your task is to write a program that:1) Prompts the user for the number of work weeks available (integer);2) Prompts the user for the name of the input file (string);3) Prompts the user for the name of the output file (string);4) Reads the available projects from the input file;5) Dolves the corresponding knapsack problem, without repetition of items; and6) Writes to the output file a summary of the results, including the expected profit and a list of the bestprojects for the company to undertake.Here is a sample session with the program:Enter the number of available employee work weeks: 10Enter the name of input file: KnapsackData1.txtEnter the name of output file: Output1.txtNumber of projects = 4DoneFor the above example, here is the output that should be written to Output1.txt:Number of projects available: 4Available employee work weeks: 10Number of projects chosen: 2Number of projectsTotal profit: 46Project0 6 30Project2 4 16The file KnapsackData2.txt, contains one thousand prospective projects. Your program should also be able to handle this larger problem as well. The corresponding output file,WardOutput2.txt, is below.With a thousand prospective projects to consider, it will be impossible for your program to finish in areasonable amount of time if it uses a "brute-force search" that explicitly considers every possiblecombination of projects. You are required to use a dynamic programming approach to this problem.WardOutput2.txt:Number of projects available: 1000Available employee work weeks: 100Number of projects chosen: 66Total profit: 16096Project15 2 236Project73 3 397Project90 2 302Project114 1 139Project117 1 158Project153 3 354Project161 2 344Project181 1 140Project211 1 191Project213 2 268Project214 2 386Project254 1 170Project257 4 427Project274 1 148Project275 1 212Project281 2 414Project290 1 215Project306 2 455Project334 3 339Project346 2 215Project356 3 337Project363 1 159Project377 1 105Project389 1 142Project397 1 321Project399 1 351Project407 3 340Project414 1 266Project431 1 114Project435 3 382Project446 1 139Project452 1 127Project456 1 229Project461 1 319Project478 1 158Project482 2 273Project492 1 142Project525 1 144Project531 1 382Project574 1 170Project594 1 125Project636 2 345Project644 1 169Project668 1 191Project676 1 117Project684 1 143Project689 1 108Project690 1 216Project713 1 367Project724 1 127Project729 2 239Project738 1 252Project779 1 115Project791 1 110Project818 2 434Project820 1 222Project830 1 179Project888 3 381Project934 3 461Project939 3 358Project951 1 165Project959 2 351Project962 1 316Project967 1 191Project984 1 117Project997 1 187

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period. b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits. c. Do all of these samples meet the contract specifications? Explain your answer. d. What trend do you observe based on the data? What could cause this trend?"

Determine the degrees of superheat of steam at 101.325 kPa and 170°C. a. 50°C b. 70°C c. 60°C d. 80°C

IN JAVA,Knapsack ProblemThe file KnapsackData1.txt and KnapsackData2.txt are sample input filesfor the following Knapsack Problem that you will solve.KnapsackData1.txt contains a list of four prospective projects for the upcoming year for a particularcompany:Project0 6 30Project1 3 14Project2 4 16Project3 2 9Each line in the file provides three pieces of information:1) String: The name of the project;2) Integer: The amount of employee labor that will be demanded by the project, measured in work weeks;3) Integer: The net profit that the company can expect from engaging in the project, measured in thousandsof dollars.Your task is to write a program that:1) Prompts the user for the number of work weeks available (integer);2) Prompts the user for the name of the input file (string);3) Prompts the user for the name of the output file (string);4) Reads the available projects from the input file;5) Dolves the corresponding knapsack problem, without repetition of items; and6) Writes to the output file a summary of the results, including the expected profit and a list of the bestprojects for the company to undertake.Here is a sample session with the program:Enter the number of available employee work weeks: 10Enter the name of input file: KnapsackData1.txtEnter the name of output file: Output1.txtNumber of projects = 4DoneFor the above example, here is the output that should be written to Output1.txt:Number of projects available: 4Available employee work weeks: 10Number of projects chosen: 2Number of projectsTotal profit: 46Project0 6 30Project2 4 16The file KnapsackData2.txt, contains one thousand prospective projects. Your program should also be able to handle this larger problem as well. The corresponding output file,WardOutput2.txt, is below.With a thousand prospective projects to consider, it will be impossible for your program to finish in areasonable amount of time if it uses a "brute-force search" that explicitly considers every possiblecombination of projects. You are required to use a dynamic programming approach to this problem.WardOutput2.txt:Number of projects available: 1000Available employee work weeks: 100Number of projects chosen: 66Total profit: 16096Project15 2 236Project73 3 397Project90 2 302Project114 1 139Project117 1 158Project153 3 354Project161 2 344Project181 1 140Project211 1 191Project213 2 268Project214 2 386Project254 1 170Project257 4 427Project274 1 148Project275 1 212Project281 2 414Project290 1 215Project306 2 455Project334 3 339Project346 2 215Project356 3 337Project363 1 159Project377 1 105Project389 1 142Project397 1 321Project399 1 351Project407 3 340Project414 1 266Project431 1 114Project435 3 382Project446 1 139Project452 1 127Project456 1 229Project461 1 319Project478 1 158Project482 2 273Project492 1 142Project525 1 144Project531 1 382Project574 1 170Project594 1 125Project636 2 345Project644 1 169Project668 1 191Project676 1 117Project684 1 143Project689 1 108Project690 1 216Project713 1 367Project724 1 127Project729 2 239Project738 1 252Project779 1 115Project791 1 110Project818 2 434Project820 1 222Project830 1 179Project888 3 381Project934 3 461Project939 3 358Project951 1 165Project959 2 351Project962 1 316Project967 1 191Project984 1 117Project997 1 187

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period. b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits. c. Do all of these samples meet the contract specifications? Explain your answer. d. What trend do you observe based on the data? What could cause this trend?"

Determine the degrees of superheat of steam at 101.325 kPa and 170°C. a. 50°C b. 70°C c. 60°C d. 80°C

The **annual quantity** of **welded** assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the **total costs** for each method and then equate them.

For the manual arc welding cell:

Labor **cost** per hour = (welder's **hourly rate** x arc-on time) + (fitter's **hourly rate** x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25

Labor cost per **welded** assembly = labor cost per hour x cycle time per assembly / 60 = $11.25 x 15.4 / 60 = $2.89

Overhead cost per **welded** assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68

Total cost per **welded** assembly = labor cost per **welded** assembly + overhead cost per **welded** assembly = $2.89 + $4.68 = $7.57

Total cost per hour = total cost per **welded** assembly x production rate = $7.57 x 8 = $60.56

Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 = $121,120

For the robotic arc welding cell:

Labor cost per hour = fitter's **hourly rate** x fitter's participation in the cycle = $25 x 0.62 = $15.50

Labor cost per **welded** assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 = $3.97

Overhead cost per **welded** assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 + $3,500 / (8 x 2,000) = $3.80 + $0.22 = $4.02

Total cost per **welded** assembly = labor cost per **welded** assembly + overhead cost per **welded** assembly + (installed cost / (production rate x service life)) = $3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 + $4.02 + $6,208.33 = $14.19

Total cost per hour = total cost per **welded** assembly x production rate = $14.19 x 8 = $113.52

Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 = $227,040

To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the **annual****quantity** of **welded** assemblies:

$121,120 + x($7.57) = $227,040 + x($14.19)

$7.57x - $14.19x = $227,040 - $121,120

$-6.62x = $105,920

x = $105,920 / $6.62

x = 15,982.7

Therefore, the **annual****quantity** of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

Learn more about **hourly rate **here brainly.com/question/29335545

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b. A rigid bar does not bend regardless of the loads acting upon it.

c. A rigid bar deforms when experiencing applied loads.

d. A rigid bar is unable to translate or rotate about a support.

e. A rigid bar represents an object that does not experience deformation of any kind.

**Answer:**

option b and E are true

**Explanation:**

A lever is an example of a rigid bar that can rotate around a given point. In a rigid material, the existing distance does not change whenever any load is placed on it. In such a material, there can be no deformation whatsoever. Wit this explanation in mind:

option a is incorrect, given that we already learnt that no deformation of any kind happens in a rigid bar.

__ option b is true.__ A rigid bar remains unchanged regardless of the load that it carries.

option c is incorrect, a rigid bar does not deform with loads on it

option d is incorrect. A lever is a type of rigid bar, a rigid bar can rotate around a support.

__ option e is true.__ A rigid bar would not experience any deformation whatsoever.

**Solution :**

**i. Slip plane (1 1 0)**

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

τ = σ cos Φ cos λ

∴

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

τ = σ cos Φ cos λ

∴

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

τ = σ cos Φ cos λ

∴

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

τ = σ cos Φ cos λ

∴

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

**Answer:**

**tensile stress at yield = 254 MPa**

**ultimate stress = 477 MPa**

**average stress = 892 MPa**

**Explanation:**

**Given data in question**

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

**to find out **

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

**solution**

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield = yield load / area

tensile stress at yield = 80 ×10³ / /4 × D²

tensile stress at yield = 80 ×10³ / /4 × 0.020²

**tensile stress at yield = 254 MPa**

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³ / /4 × 0.020²

**ultimate stress = 477 MPa**

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³ / /4 × d²

average stress = 70 ×10³ / /4 × 0.010²

**average stress = 892 MPa**

**Answer:**

Average heat flux=3729.82 W/

**Explanation:**

**Answer:**

**Explanation:**

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.