Singapore is a small island nation that faces several challenges when it comes to developing new water resources and managing existing resources for long-term water security.
Singapore's water supply has always been limited by its small size, limited land area, and lack of natural water sources. These challenges have led to the development of innovative technologies and policies to ensure water security.
One of the biggest challenges facing Singapore in developing new water resources is the country's limited land area. To overcome this challenge, Singapore has implemented a range of measures, including the development of new water sources, such as desalination plants, NEWater facilities, and rainwater harvesting systems.
In addition to developing new water resources, Singapore also faces challenges in managing its existing resources for long-term water security.
One of the biggest challenges is the limited availability of freshwater resources, which are vulnerable to pollution and overuse. Another challenge facing Singapore in managing its existing water resources is the impact of climate change. Singapore is vulnerable to rising sea levels, which could lead to saltwater intrusion and contamination of freshwater sources.
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D2. An office building for multiple tenants with 20 floors (above the main terminal) of 600m² per floor to be built. The estimated population per floor is 60 persons and the estimated up peak demand is 14%. The floor-to-floor height of the building is 3m and the design intervals is 50 seconds, the passenger transfer time is 1.4 second, door opening and closing time are 1 second and 3 second respectively, interfloor jump time is 3 second and the rated speed of the lift is 4 m/s, determine the following details of the lift system with the lift car capacity factor of 80%.
(i) Round Trip Time (RTT);
(ii) Number of lift car required;
(iii) Contract capacity (CC) of each lift car;
(iv) Actual Up peak Interval (UPPINT);
(v) Actual Up peak Handling capacity (UPPHC);
(vi) Actual Percentage of population (%POP)
D2. An office building for multiple tenants with 20 floors (above the main terminal) of 600m² per floor to be built. The estimated population per floor is 60 persons and the estimated up peak demand is 14%.
The floor-to-floor height of the building is 3m and the design intervals are 50 seconds, the passenger transfer time is 1.4 seconds, door opening and closing time are 1 second and 3 seconds respectively, inter floor jump time is 3 seconds and the rated speed of the lift is 4 m/s, determine the following details of the lift system with the lift car capacity factor of 80%.
(i) Round Trip Time (RTT); The round trip time is the time taken by a lift to travel from the ground floor to the highest floor of the building and return to the ground floor. The RTT can be calculated using the formula: RTT = 2 × Single Distance/Speed where Single Distance = (Number of floors – 1) × Floor-to-Floor height Single Distance = (20 - 1) × 3 = 57m Speed = 4m/s RTT = 2 × 57/4 = 28.5 seconds(ii) Number of lift car required.
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NPDC plans to develop a field in the Niger Delta. The field to be developed has the potential of 80,000,000 (80MM) barrels of oil. It is estimated that development will take about 4 years, with production beginning at the end of the third year. Facilities will be constructed to handle 20,000 barrels of oil per day. It is estimated that after production reaches the limit of the facilities at the end of the sixth year, it will continue to produce at that rate for 6 years before beginning to decline at a constant percentage per year. This constant production rate is known as the plateau rate. The economic limit has been determined to be 600 barrels of oil per day. Determine A forecast of annual production rate and cumulative production over the entire life of this field.
The field in Niger Delta, developed by NPDC has an estimated potential of 80,000,000 (80MM) barrels of oil. The development is estimated to take about 4 years. The production is assumed to start at the end of the third year, and facilities will be constructed to handle 20,000 barrels of oil per day.
The economic limit is determined to be 600 barrels of oil per day. Determine a forecast of the annual production rate and cumulative production over the entire life of this field. Annual production rate: Annual production rate = 20,000 barrels of oil per day × 365 days Annual production rate = 7,300,000 barrels of oil per year.
At the end of the sixth year, the facility limit is expected to be reached, and the production will continue at a constant rate of 20,000 barrels of oil per day until the 12th year. Cumulative production: Production from year 0 to year 6 = 7,300,000 barrels of oil per year × 6 years.
Production from year 0 to year 6 = 43,800,000 barrels of oil Cumulative production from year 7 to year 12:Annual production rate = 20,000 barrels of oil per day × 365 days Annual production rate = 7,300,000 barrels of oil per year Cumulative production from year 7 to year 12 = 7,300,000 barrels of oil per year × 6 years Cumulative production from year 7 to year 12 = 43,800,000 barrels of oil.
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Jane is in a position to specify several purchases that will benefit a firm in which she has a substantial financial investment.
This case does not describe a conflict of interest situation.
She can escape the conflict of interest situation as long as she makes the right decision from a professional standpoint.
She can escape the conflict of interest situation if she "recuses" (or dismiss) herself from the decision-making
She can escape the conflict of interest situation if she discloses her private interest in the bolt-manufacturing firm.
Conflict of interest is a situation where an individual has competing interests or loyalties. The individual must balance the competing interests, and choosing one over the other may result in harm or damage to either interest.
Jane is in a position to specify several purchases that will benefit a firm in which she has a substantial financial investment. Though this case does not describe a conflict of interest situation, it is essential to know how to handle one if the situation arises.
She can escape the conflict of interest situation as long as she makes the right decision from a professional standpoint. In other words, Jane should put her professional duties above her financial interest to avoid a conflict of interest situation.
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A cylindrical specimen of a brass alloy having a length of 92 mm (3.622 in.) must elongate only 4.60 mm (0.1811 in.) when a tensile load of 92000 N(20680lb f ) is applied. Under these circumstances what must be the radius of the specimen? Consider this brass alloy to have the stress-strain behavior shown in the Animated Figure 6.12. mm
The radius of the brass alloy specimen is determined to be 6.890 mm.
Let's begin by applying the formula:
Stress = Force / Area
and
Strain = Extension / Original Length
The given data:
Original length of cylindrical specimen of brass alloy = 92 mm (3.622 in.)
Force applied (F) = 92000 N
Extension (e) = 4.60 mm (0.1811 in.)
Stress-strain curve is given in the figure below:
Animated Figure 6.12 [Credit: T. Ahrens, Essentials of Materials Science and Engineering, 2nd edition (Prentice Hall, Upper Saddle River, NJ, 2008)]
To determine the radius of the specimen, we need to find the cross-sectional area of the cylindrical specimen of brass alloy. Since the material is isotropic (properties in all directions are same), the stress will be the same in the radial direction as it is in the longitudinal direction. So, the stress-strain curve remains the same for the radial and longitudinal directions.
Hence, stress in the radial direction,
σ = Force / Area = F / [πr²]
Where, r is the radius of the cylindrical specimen.
The elongation is given by,
Strain = Extension / Original Length = e / l
Substituting values in the above formula, we get;
σ/E = (ln(ε+1))
(σ/E) = ln (1+ε)
So,
σ = E * ln (1+ε)
Where,
E is the modulus of elasticity (Young's modulus) of the brass alloy. From the given figure;
Young's modulus (E) = 103 GPa = 103 x 109 N/m² = 103 x 103 N/mm²
When strain is 0.006, ln (1+ε) is 0.005987.
So,
σ = 103 x 103 x 0.005987 = 618.361 N/mm²
Area of cross-section,
A = F / σ = 92000 / 618.361 = 148.70 mm²
Area of cross-section of cylindrical specimen,
A = πr²
πr² = 148.70 mm²
r² = 148.70 / π
r² = 47.35mm²
r = 6.890 mm
The radius of the specimen is 6.890 mm.
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Select the appropriate uses of each of the following construction equipment specifically for this Project:
A. 1. Ten-Wheeler Dump trucks
2. Bulldozers
3. Wheeled Pay Loaders
4. Motor Graders
5. Backhoes
6. Wheel Tractor Scrapers,
7. Compactors,
8. Concrete Mixers.
9. Wheeled Tractor Scraper
a) To load accumulated or piles rocks, soil, and sand to the s. Dump trucks
b) To delivery ready mixed concrete to the project site
c) To carry the rocks and soil in larger quantities from one site to another site or to the dump yard
d) To level the soil surface in preparation for laying out of rebars and metal forms along the sides
e) to remove the top soil layers consists of rocks, soil to clear the site for road construction
f) To excavate trenches for laying out of drainage culverts
8) To compact the leveled top soil in preparation for laying out of the rebars and metal form on the sides.
h) To remove rocks, sand, and soil dumped and load in the dump trucks
i) To flatten soil surfaces by scraping
Here are the appropriate uses of each of the following construction equipment for this project:1. Ten-Wheeler Dump trucks a) To load accumulated or piles rocks, soil, and sand to the dump yard. b) To carry the rocks and soil in larger quantities from one site to another site or to the dump yard.
c) To remove rocks, sand, and soil dumped and load in the dump trucks.2. Bulldozers d) To level the soil surface in preparation for laying out of rebars and metal forms along the sides. e) To remove the top soil layers consisting of rocks, soil to clear the site for road construction.3. Wheeled Pay Loaders f) To excavate trenches for laying out of drainage culverts.
4. Motor Graders i) To flatten soil surfaces by scraping.5. Backhoes f) To excavate trenches for laying out of drainage culverts.6. Wheel Tractor Scrapers a) To load accumulated or piles rocks, soil, and sand to the dump yard. b) To carry the rocks and soil in larger quantities from one site to another site or to the dump yard.7. Compactors h) To remove rocks, sand, and soil dumped and load in the dump trucks.
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NB 1: This assignment is designed to (a) develop your appreciation of the principles relating to ground stress distribution and immediate settlement, and (b) improve your skills in the areas of conceptual modelling of practical engineering problems, use of manual computations and software simulations for problem solving, working as part of a team, and effectively communicating the results.
N.B. 2: This assignment is to be completed by students working in teams of three (students should organize their groups).
A 5 m radius and 12 m height water storage tank is to be constructed close to one of the team members' residence. Complete the tasks below related to this planned construction.
Task A:
(i) Determine the factor of safety against a bearing capacity failure
(ii) Plot the ground stress distribution beneath the foundation of the tank by calculating the stress increase using two different empirical methods
(iii) Calculate immediate settlement beneath the foundation of the tank
Task A:(i) The factor of safety against a bearing capacity failure can be determined as follows:Nγ Df + 0.4γ BNq ≤ qf σV = 1.5γBNγfNq+ 0.4γDNγq ≤ σallwhere,
Nγ = Bearing capacity factor = 34.9
Nq = Bearing capacity factor = 28.1
γf = Factor of safety = 2.5
γB = Soil density = 19.2 kN/m3
γD = Density of water = 9.81 kN/m3
Df = Foundation depth = 1.5 m
B = Foundation width = 3 m
qf = Applied load per unit area = 1.5 × 20 × 1.5 = 45 kN/m2σall = Allowable soil pressure = 400 kN/m2By substituting the values of the different factors in the formula, we get:
Nγ Df + 0.4γ BNq ≤ qf34.9 × 1.5 + 0.4 × 19.2 × 28.1 ≤ 45Hence the factor of safety against a bearing capacity failure is greater than 1.5, which is the recommended value.(ii) The ground stress distribution beneath the foundation of the tank can be determined using two empirical methods as follows:
(a) Boussinesq’s TheoryAccording to Boussinesq’s theory, the vertical stress at a point directly beneath a point load on the surface of a half-space can be calculated asσz = Q / [(1 + v) π r2 (z2 + r2)]3/2where,
Q = Point load = 45 kN/m2
v = Poisson’s ratio = 0.3
r = Radius of the foundation = 5 m
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Advances in architectural achievements thrived during the Aegean and Greek eras, creating visual inspirations still used today. Investigate the progress of architecture and describe the development of architecture in this time period.
Paragraph 1: Aegean Architecture (Mycenean and/or Minoan).
Paragraph 2: Greek Architecture (Archaic, Classical, and/or Hellenistic)
Can you please answer these two parts by what the type of architecture is known for, in all aspects, and using a photo from each topic to analyze them?
Advances in architectural achievements thrived during the Aegean and Greek eras, creating visual inspirations still used today. Let's investigate the progress of architecture and describe the development of architecture in this time period.
Aegean Architecture: The Aegean era, which includes the Minoan and Mycenaean cultures, thrived between 2500 BC and 1200 BC, significantly before the Greek era, and is believed to have laid the foundation for Greek culture. In general, the Aegean architecture is characterized by a preference for curved lines, rounded corners, and bright colors, with a strong emphasis on nature and its surroundings.
Minoan palaces, like the one depicted below, were the most impressive structures of the time. The Knossos palace in Crete is an example of one such palace, which boasted a labyrinthine layout that was intended to be both practical and aesthetically pleasing, with various colors and patterns. It had a number of fortified walls and towers, as well as a large central space, the megaron, which was the palace's focal point.
Greek Architecture: Greek architecture is often split into three categories: Archaic, Classical, and Hellenistic. This time period lasted from approximately 700 BC to 30 BC, and it was marked by significant technological advances and the development of complex designs. The Altar of Pergamon, an ancient Greek structure from the Hellenistic era, is depicted below. The structure had a unique design and was made of white marble.
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In a direct shear test using a shear box of 75 x 75 mm in plan. The shear machine is designed to operate with the lever arm assembly arranged to multiply a hangar load by 8. Compute a normal stress (kPa) on the sample when when the shear force is 384 N ( 1 dp). Use gravitational acceleration of 9.81 m/s2.
The shear box of 75 × 75 mm in plan and the lever arm assembly is arranged to multiply a hangar load by 8.The shear force is 384 N (1 dp) and the gravitational acceleration is 9.81 m/s².
To calculate the normal stress, we can use the below formula, Normal stress, σn = P/A Here, P = Force acting on the sample A = Area of the sample In the given problem, Area of the sample = 75 × 75 mm² = 5.625 × 10⁴ mm²Force acting on the sample = 384 N (given) = 384 × 10⁶ N/m².
Now we will find the hangar load by dividing the force acting on the sample by the lever arm arrangement .So,Hangar load = (384 × 10⁶) / 8Hangar load = 48 × 10⁶ N/m²Now,Normal stress, σn = P/A Normal stress, σn = (48 × 10⁶) / 5.625 × 10⁴Normal stress, σn = 853.33 kPa (approx.)Therefore, the normal stress on the sample is 853.33 kPa (approx.) when the shear force is 384 N.
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h= 300mm b =190mm t = 4mm L/h = 12.5
Question: Consider self weight only. Draw the stress and strain distributions at midspan.
[tex]Given data:h = 300 mGivenm b = 190 mm t = 4 mm L/h = 12.5[/tex]Consider self-weight onlyThe maximum stress in a beam will occur either at the supports or at the center of the beam if it's a simply supported beam.
Since it is a simply supported beam, therefore, the maximum stress will occur at midspan.The distribution of stress and strain in a beam under bending depends on the bending moment distribution and the distribution of the cross-section area from the neutral axis. The neutral axis is that imaginary.
Line within the cross-section of the beam which undergoes no change in length or volume during the deformation of the beam. The stress is maximum on the top and minimum at the bottom, and the strain is maximum at the outermost fibers and minimum at the neutral axis.Now,Let's calculate the values of various quantities.
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Problem 1 An animal farm generates 35 tonnes of organic waste (animal manure) on a daily basis. The transportation of this waste costs approximately JD 0.55/tonne/km. There are three alternatives available for the management and disposal of this waste: • Option (1): Disposal in an open dumpsite which is nearly 18 km far from the farm. The average tipping fee at the disposal site's gate is JD 6/tonne. • Option (2): Giving away to farmers (as a fertilizer). This requires transporting the waste to another nearby town (95 km away). • Option (3): Onsite composting, which requires the construction and operation of a small onsite composting plant. The average cost (capital and operating) of composting is roughly JD 54/tonne while the final product (compost) can be sold for JD 25/tonne. The average greenhouse gas (GHG) emission factors are as follows: Open disposal = 0.25 g CO2 per g waste, direct fertilizer application = 0.13 g CO2 per g waste; composting = 0.09 g CO2 per g waste; and transportation = 56 g CO₂/ton waste/km. Answer the following questions: a. Rank the three alternatives in terms of their environmental-friendliness and explain your answer. b. From a pure economic perspective, which option should be selected and why? Show your calculations if any. c. From a multi-criteria standpoint (economic, environmental, societal), which option should be selected? Explain the impact of the selected option on all three criteria.
Option 3, Onsite composting should be ranked first because the average greenhouse gas (GHG) emission factor is lowest at 0.09 g CO2 per g waste compared to the other alternatives.
Option 2, Giving away to farmers should be ranked second because direct fertilizer application has an average greenhouse gas (GHG) emission factor of 0.13 g CO2 per g waste.
Option 1, Disposal in an open dumpsite which is ranked third as it has the highest average greenhouse gas (GHG) emission factor of 0.25 g CO2 per g waste.
b. From a pure economic perspective, .Option 2, Giving away to farmers (as a fertilizer) should be selected from a pure economic perspective because it has the lowest total cost of [tex]JD 0.55 * 95 = JD 52.25 per tonne[/tex].
Thus, the pure economic perspective option should be option 2.
c. From a multi-criteria standpoint (economic, environmental, societal), Option 3, Onsite composting should be selected based on a multi-criteria standpoint (economic, environmental, societal) because it has a low GHG emission factor of 0.09 g CO2 per g waste, reduces the cost and environmental effects of transportation, and creates jobs.
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1) Which of the following provisions would you expect to see in virtually all confidentiality agreements obtained by privately held target companies?
a. Employee non-solicitation provision and drag-along provision
b. Use restriction and disclosure restriction
c. Standstill provision and residuals clause
d. All of the above
2)
In an antitrust-sensitive acquisition, the target company may allocate some of the antitrust risk to the buyer by insisting that the definitive acquisition agreement include:
a. A hell or high water (HOHW) antitrust covenant
b. A reverse break-up fee payable by the buyer upon termination of the acquisition agreement because of an antitrust challenge
c. A covenant on the part of the buyer to litigate any antitrust challenge by the antitrust authorities
d. All of the above
Use restriction and disclosure restriction are the two provisions that would be expected to be seen in almost all confidentiality agreements obtained by privately held target companies. A confidentiality agreement is a legally binding agreement between two or more parties in which at least one of the parties agrees not to disclose certain confidential information to outsiders or the public.1) The correct option is b.
2) The correct option is d. A hell or high water (HOHW) antitrust covenant, a reverse break-up fee payable by the buyer upon termination of the acquisition agreement because of an antitrust challenge, and a covenant on the part of the buyer to litigate any antitrust challenge by the antitrust authorities are all included in the definitive acquisition agreement in an antitrust-sensitive acquisition.
When purchasing a company, a buyer takes on a variety of risks, including antitrust risk. As a result, the target company may shift some of the antitrust risk to the buyer by insisting that the definitive acquisition agreement include antitrust provisions.
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walls are identical. The internal temperature of the building is approximately 18°C and outside temperature is approximately -2°C.The wall has the following specifications: A 30m² area of wall forms one side of a simple 4 walled building where all four
Internal surface resistance 0.120m²K/W
External surface resistance 0.060m²K/W
Cavity resistance 0.180m²K/W
15mm plaster I-value 0.450W/mK 100mm block l-value 0.216W/mK
30mm partial cavity fill l-value 0.036W/mK
102mm brick l-value 0.840W/mK
a. Calculate the U-value and thus the rate of fabric heat loss across the wall.
b. The room is 2.25m high and has two air changes per hour. The volumetric heat capacity of the air is 1300J/m³K.Calculate the rate of ventilation heat loss and thus the total heat loss from the room. You may ignore the loss through the floor and roof. c. What surface area of radiators would be needed to compensate for the above conditions if the radiators output is 400W/m²
d. If the total heat gain received by this room amounts to 10800MJ per heating quarter, calculate the quarterly heating bill, based upon maintaining the thermal comfort within the room if replacement heat costs 2.5p/MJ.
a. Calculation of U-value and Fabric heat loss rate across the wall U-value is the measure of the rate at which heat can be transmitted through the wall. It is determined as:
Rtotal = Rinternal + Rexternal + Rcavity + Rbrick+ Plaster I-value + Partial cavity fill I-valueU-value is the inverse of RtotalU-value = 1 / RtotalRate of heat loss per unit area of the wall can be determined as:[tex]Q = U × A × (Tinternal - Texternal) = 0.285 kW[/tex]
The U-value is[tex]0.285/16.67=0.0171kW/m2K[/tex]which is the rate of fabric heat loss across the wall. (where A= 30 m2)Hence the U-value and fabric heat loss rate across the wall is 0.0171 kW/m²K and 0.285 kW/m² respectively.
b. total heat capacity =[tex]67.5 x 1.3 = 87.75 kJ/K[/tex]For 2 air changes per hour, the air change per minute will be 2 / 60 = 1/30Therefore, the rate of ventilation heat loss can be determined as:[tex]Hv = Qv × Δt × V[/tex]For air changes of 2 per hour, [tex]Δt = 18 - (-2) = 20HV = (1/30) × 87.75 × 20 = 58.5[/tex] The total heat loss from the room is given as
:[tex]HTotal = Qf + Hv,Qf = 0.285 kW/m²[/tex] Hence, [tex]HTotal = 0.285 × 30 + 58.5 = 66.45 kW[/tex]
c. Aradiator = Hsupplied / Pradiator = [tex]66.45 / 400 = 0.166 m²[/tex]Hence, the surface area of radiators needed is 0.166 m².
d. the total cost of replacement heat per quarter is:
Total heat cost = Heat gain × cost per unit energy =[tex]10800 x 2.5 / 100 = £270[/tex]Hence, the quarterly heating bill would be £270 if replacement heat costs 2.5p/MJ.
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Your team of consultants has been hired by the city of Laramie to help them design an effective bus system. The city has given you their planning map as a starting point, containing major streets (see Figure 1). There are six zones, shaded in gray in the figure (North Taramie, South Laramie, East Laramie, West Laramie, UW, and Downtown). They also give you files containing node coordinates and link information but they have left the OD matrix blank because their model is out of date. You are also given zone information from the census (Table 1), results from a recent travel survey (Table 2) and a table of friction factors (Table 3).
The city's budget allows them to operate three buses, and they want to know how to do so in the way which is the most helpful to Laramie citizens, measured according to the total ridership. To satisfy your contract with the city, your team has to accomplish the following tasks:
1. Run the four-step model to identify the travel time on each roadway link during the AM peak, off-peak, and PM peak. For this base case, there is no bus system, so you should skip mode choice and assume everyone will drive.
2. Identify the routes each bus will take during the AM peak, off-peak, and PM peak periods. Each route must be a loop, and you can describe it by the node numbers it passes. More than one bus can use the same route.
3. Calculate the total bus ridership for your route choices (described below). You might need the following information as well for the first task:
• As described in class, we only consider work and shopping trips. (For this assignment you might think of "work" trips to LW including students going to class) Each work production results in one trip from home to work in the AM peak, and one from work to home in the PM peak. Each shopping production results in one trip from home to shopping and a return trip from shopping to home during the off-peak period.
The equation for work attractions to a zone is Aw-125w, where w' is the number of workplaces or schools in that Zone.
. The equation for shopping attractions to a zone is As 10-T/100 where I is the average income in that zone, and T is the sales tax receipts from that zone (in thousands of dollars).
• Assume that the travel times will not change after the bus system is in place (that is, you do not have to redo route choice after.
• Separate network files are provided for the AM/PM peak periods and the off-peak because the latter is longer (and therefore roadway "capacity" is higher).
For the third task, you calculate ridership using the following procedure:
1. Identify the OD pairs which might possibly use the bus - assume that if there is no bus route directly connecting an origin to a destination, nobody from that OD pair will travel by bus. That is to say, we assume nobody will transfer buses, take a bus to UW and walk downtown, etc
2. For these OD pairs, calculate the travel time by driving (shortest path from origin to destination using the travel times found in the first task) and the travel time on the bus (take the travel time on the bus links from that origin to that destination and increase it by 20% to account for the bus stopping and driving more slowly)
3. Calculate the "frequency" of each bus as the reciprocal of the total travel time on its route. (e.g., if the bus takes 50 minutes to complete its loop, its frequency is 1/50)
4. For each OD pair using the bus:
(a) Calculate the "total frequency as the sum of the frequencies of each bus connecting that OD pair. (e.g., if there two buses connecting that origin and destination, one of which comes every
50 minutes and one of which comes every 20 minutes, the total frequency is 1/50+ 1/20) (b) Calculate the utility of the driving as Udr-Tar where Tar is the travel time driving.
(c) Calculate the utility of taking the bus as Ubus-1.25Tbus+ Fbus where Tous is the travel time by bus and Fbus is the total frequency.
(d) Find the bus ridership from this OD pair using these utilities.
5. Add the ridership from each OD pair to get the total ridership
Tum in your answers to the above tasks, along with any supporting documents
In order to design an effective bus system, the consultants have to perform the following tasks :Run the four-step model to identify the travel time on each roadway link during the AM peak, off-peak, and PM peak.
Calculate the total bus ridership for the chosen routes.1. Running the four-step model The first step is to use the four-step model to find the travel time on each roadway link during the AM peak, off-peak, and PM peak. Since there is no bus system in place, the consultants should assume that everyone will drive. 2. Identifying routes for busesThe second step is to identify the routes each bus will take during the AM peak, off-peak, and PM peak periods. It is important to note that each route must be a loop, and more than one bus can use the same route.3. Calculating total bus ridership The third step is to calculate the total bus ridership for the chosen routes.
For these OD pairs, calculate the travel time by driving and the travel time on the bus. Calculate the frequency of each bus as the reciprocal of the total travel time on its route. For each OD pair using the bus, calculate the total frequency as the sum of the frequencies of each bus connecting that OD pair. Calculate the utility of the driving and the utility of taking the bus. Find the bus ridership from this OD pair using these utilities. Add the ridership from each OD pair to get the total ridership.
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TRUE OR FALSE
16. Combination load factor for permanent load could be less than 1.0.
17. AS1170.0 covers earthquake actions.
18. Loads due to non-structural components shall be transferred to the foundation.
19. Value of uplift force for designing anchors between wall and roof for a house in Darwin could be zero.
20. Maximum slope of an escarpment for calculating wind pressure is the steepest average slope within the top H/2 (half of height) in any direction of escarpment.
21. Pressure on area of elevation for a single-story building in wind class N2 with hip roof, width of 10 m and roof pitch of 20 degrees is 0.61 kPa.
22. Maximum slope of an escarpment for calculating wind pressure is the steepest average slope within the top H/2 (half of height) in the wind direction.
23. Ultimate strength pressure for wind class N1 used for designing a wall stud is 0.62.
16. True. Combination load factor for permanent load could be less than 1.0 because the dead load is always present and it contributes to the gravity forces.
17. True. AS1170.0 covers earthquake actions as it is the basic of the Loading Code of Australia.
18. True. Loads due to non-structural components shall be transferred to the foundation because non-structural components can come off or overturn under strong winds or earthquake forces.
19. True. The value of uplift force for designing anchors between wall and roof for a house in Darwin could be zero because Darwin is located in a wind region that is not prone to cyclones
.20. True. The maximum slope of an escarpment for calculating wind pressure is the steepest average slope within the top H/2 (half of the height) in any direction of the escarpment.
21. True. The pressure on the area of elevation for a single-story building in wind class N2 with hip roof, width of 10 m and roof pitch of 20 degrees is 0.61 kP.
22. True. The maximum slope of an escarpment for calculating wind pressure is the steepest average slope within the top H/2 (half of height) in the wind direction.
23. False. Ultimate strength pressure for wind class N1 used for designing a wall stud is not 0.62 as the ultimate strength pressure varies with the wind class.
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Discuss the concept of leadership and also illustrate the styles of leadership with practical examples, being a leader what kind of style should be properly implemenetd at work place to satisfy your subordinates that is directly influence on your organisation objective
Leadership is the act of directing and guiding people or an organization towards a particular goal or objective. It involves creating a vision, inspiring others, and enabling them to achieve their full potential. In order to achieve this, a leader must have a variety of skills and qualities that enable him or her to lead effectively.
There are different styles of leadership that can be used by leaders to achieve their goals and objectives. These styles include:
1. Autocratic Leadership: This is a leadership style in which the leader has complete control over the decisions and actions of his or her subordinates. The leader makes all the decisions and does not consult or seek input from his or her subordinates. This style is often used in situations where quick decisions need to be made or in situations where the leader has more expertise than his or her subordinates. An example of autocratic leadership is a military general.
2. Democratic Leadership: This is a leadership style in which the leader involves his or her subordinates in decision-making. The leader consults with his or her subordinates and seeks their input before making a decision. This style is often used in situations where the leader does not have all the information or expertise required to make a decision on his or her own. An example of democratic leadership is a group leader in a project team.
3. Laissez-Faire Leadership: This is a leadership style in which the leader provides minimal guidance or direction to his or her subordinates. The leader gives his or her subordinates the freedom to make their own decisions and to work independently. This style is often used in situations where the subordinates are highly skilled and have a high degree of expertise in their area of work. An example of laissez-faire leadership is a manager of a research team.
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An electrical utility is experiencing a sharp power demand that continues to grow at a high rate in a certain local area. Two alternatives are under consideration. Each is designed to provide enough capacity during the next 25 years, and both will consume the same amount of fuel, so fuel cost is not considered in the analysis.
Alternative A. Increase the generating capacity now so that the ultimate demand can be met without additional expenditures later. An investment of $29 million would be required, and it is estimated that this plant facility would be in service for 25 years and have a salvage value of $0.8 million. The annual operating and maintenance costs (including income taxes) would be $0.6 million. Alternative B. Spend $13 million now and follow this expenditure with future additions during the 10th year and the 15th year. These additions would cost $17 million and $10 million, respectively. The facility would be sold 25 years from now with a salvage value of $1.2 million. The annual operating and maintenance costs (including income taxes) will be $250,000 initially and will increase to $0.35 million after the second addition (from the 11th year to the 15th year) and to $0.45 million during the final 10 years. (Assume that these costs begin one year subsequent to the actual addition.) On the basis of the present-worth criterion, if the firm uses 14% as a MARR, which alternative should be undertaken? Note:Adopt incremental cost approach.
Based on the present-worth criterion and a 14% MARR, Alternative A should be undertaken as it has a lower present worth compared to Alternative B.
Incremental To determine which alternative should be undertaken based on the present-worth criterion and a 14% MARR, we need to calculate the present worth of each alternative's costs.
For Alternative A:
Initial investment: -$29 million
Annual operating and maintenance costs: -$0.6 million per year for 25 years
Salvage value: +$0.8 million
For Alternative B:
Initial investment: -$13 million
Additional investments: -$17 million in the 10th year, -$10 million in the 15th year
Annual operating and maintenance costs: -$0.25 million for the first 10 years, -$0.35 million from the 11th to the 15th year, -$0.45 million for the final 10 years
Salvage value: +$1.2 million
Using the incremental cost approach, we calculate the present worth of costs for each alternative. We consider the difference in costs between the alternatives over their respective lifetimes.
For Alternative A:
Present worth = -29 + (0.6 * (1 - (1 + 0.14)^-25) / 0.14) + 0.8 / (1 + 0.14)^25
For Alternative B:
Present worth = -13 + (0.25 * (1 - (1 + 0.14)^-10) / 0.14) + (0.35 * (1 - (1 + 0.14)^-5) / 0.14) + (0.45 * (1 - (1 + 0.14)^-10) / 0.14) + 1.2 / (1 + 0.14)^25
Comparing the present worth values of the two alternatives, the alternative with the lower present worth should be undertaken.
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Under a standard owner-controlled insurance program (OCIP), the CM: a. Has no role in construction site safety b. Has less authority than the safety contractor c. Has authority for all construction site safety d. Directs the work of the safety contractor
A standard Owner-Controlled Insurance Program (OCIP) covers various construction risks and liabilities, including worker's compensation, third-party liability, and site liability.
This type of policy benefits the owner of the project, the general contractor, and other stakeholders involved in the project. A. The CM has no role in construction site safety This statement is incorrect because the Construction Manager (CM) plays a critical role in ensuring safety on construction sites.
The CM is responsible for monitoring and ensuring that all activities are carried out in compliance with the Occupational Safety and Health Administration (OSHA) guidelines. B. The CM has less authority than the safety contractor This statement is incorrect because the CM has more authority than the safety contractor.
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4.1 Software Interfaces |
< Format your work as a bulleted list with brief explanations
Describe the connections between this product and other software components (identified by name and version), including other applications, databases, operating systems, tools, libraries, websites, and integrated commercial components. State the purpose, formats, and contents of the messages, data, and control values exchanged between the software components. Specify the mappings of input and output data between the systems and any translations that need to be made for the data to get from one system to the other. Describe the services needed by or from external software components and the nature of the inter-component communications. Identify data that will be exchanged between or shared across software components. Specify nonfunctional requirements affecting the interface, such as service levels for responses times and frequencies, or security controls and restrictions.>
4.2 Hardware Interfaces
< Format your work as a bulleted list with brief explanations
Describe the characteristics of each interface between the software and hardware (if any) components of the system. This description might include the supported device types, the data and control interactions between the software and the hardware, and the communication protocols to be used. List the inputs and outputs, their formats, their valid values or ranges, and any timing issues developers need to be aware of. If this information is extensive, consider creating a separate interface specification document >
The connections between this product and other software components are shown below.
The characteristics of each interface between the software and hardware are :
Identify the supported device.Describe the data and control interactionsSpecify the communication4.1 Software Interfaces:
- Identify the software components and their versions that are connected to the product, including applications, databases, operating systems, tools, libraries, websites, and integrated commercial components.
- Describe the purpose, formats, and contents of the messages, data, and control values exchanged between the software components.
- Specify the mappings of input and output data between the systems and any necessary translations for data exchange.
- Describe the services required from or provided to external software components and the nature of inter-component communications.
- Identify the data that will be exchanged or shared across software components.
- Specify any nonfunctional requirements that impact the interface, such as response time and frequency service levels, security controls, and restrictions.
4.2 Hardware Interfaces:
- Describe the characteristics of each interface between the software and hardware components of the system.
- Identify the supported device types and their interactions with the software.
- Describe the data and control interactions between the software and hardware components.
- Specify the communication protocols to be used for data exchange.
- List the inputs and outputs, their formats, valid values or ranges, and any timing considerations that developers need to be aware of.
- Consider creating a separate interface specification document if the information is extensive.
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Determine the qualitative and quantitative impact of the pandemic situation to road, rail, air, and water transportation
The pandemic situation has had a major impact on the transportation sector. The transportation sector, which includes road, rail, air, and water transport, has been significantly affected by the COVID-19 pandemic. This essay will discuss the qualitative and quantitative impact of the pandemic on transportation.
Qualitative Impact
The pandemic has had a qualitative impact on transportation. The most noticeable effect of the pandemic is the decrease in the number of passengers using public transport. The fear of contracting the virus has resulted in a significant decline in the use of public transport services.
This decrease has resulted in lower revenues for public transport operators. Additionally, people have started to prefer private cars over public transport. This has led to increased traffic congestion, particularly in urban areas. Increased congestion has had a negative impact on the environment, particularly air quality.
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a) i) Briefly explain what is meant by the terms ‘grain’ and
‘grain structure’ for a metal. (2 marks)
ii) Describe why reducing grain size can increase the strength
of a pure metal. Include an
Reducing the grain size in a metal increases its strength by inhibiting the dislocation of grains and enhancing grain boundary strengthening. This is achieved through a larger surface area and increased number of grain boundaries. Consequently, smaller grain sizes contribute to improved mechanical properties and increased strength in pure metals.
i) What is meant by the terms 'grain' and 'grain structure' for a metal?
Grain: A grain refers to the crystalline lattice that represents the three-dimensional arrangement of atoms in a metal. It is formed when atoms combine and create a distinct crystal structure. The average size of grains is typically a few microns, but it can vary based on the manufacturing process. Metals cooled slowly exhibit well-defined crystal structures, while rapid cooling or cold working produces a finer grain structure.
Grain structure: Grain structure refers to the size, orientation, and arrangement of grains in a metal. The boundaries between grains are called grain boundaries, which play a crucial role in the mechanical properties of the metal. The sizes, shapes, and orientations of grains can significantly impact the metal's properties and performance.
ii) Why does reducing grain size increase the strength of a pure metal?
Reducing the grain size in a metal results in an increased total number of grains. This, in turn, leads to a larger surface area in contact with neighboring grains and an increased presence of defects at the grain boundaries. These defects, known as grain boundary strengthening, enhance the metal's strength by impeding the dislocation of grains when subjected to stress.
The dislocation of grains occurs when atoms slide and slip over one another. Having more grains inhibits these dislocations, thereby increasing the metal's strength and hardness. Consequently, reducing the grain size is effective in enhancing the strength of a pure metal.
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Engr. David is holding a permanent position as Engineer II in the Department of Public Works and Highways (DPWH). DPWH enters into a contract with Red Mole Construction (CONTRACTOR) for the construction of a school building. The Contractor wants to engage the services of Engr. David as a consultant for the project. As a form of compensation, Engr. David will be paid a percentage of the contract amount of the project.
Will the said engagement be violative of the Code of Ethics of Civil Engineers and existing laws, rules and regulations?
Under the Code of Ethics of Civil Engineers, a permanent position in a government agency prohibits a civil engineer from accepting any contract, job or work from any private entity that is related to his profession.
trade, or calling without the permission of his immediate superior. Thus, Engr. David cannot work as a consultant to Red Mole Construction without seeking permission from the Department of Public Works and Highways (DPWH) since he holds a permanent position there.
However, if the DPWH permits Engr. David to work as a consultant to Red Mole Construction, he should not receive compensation as a percentage of the contract amount of the project. The Code of Ethics of Civil Engineers stipulates that an engineer should not accept compensation on a percentage basis of the contract amount.
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which type of regenerative braking system uses an electrohydraulic system?
The hydraulic hybrid system is the type of regenerative braking system that uses an electrohydraulic system.
The electrohydraulic braking system is the type of regenerative braking system that uses an electrohydraulic system.What is Regenerative Braking System?Regenerative braking is a mechanism that enables the braking system to recover the kinetic energy produced by the wheels' motion during deceleration. Regenerative braking is used in vehicles such as cars, trucks, trains, and so on, to boost fuel economy and lower exhaust emissions. Regenerative braking recovers kinetic energy that is typically wasted when brakes are applied. This captured energy can then be used to recharge a vehicle's battery or power an electrical accessory or system.
How does the Regenerative Braking System work?Regenerative braking systems recover energy from the wheels during braking and store it in an energy storage device such as a battery or capacitor. The stored energy can then be used to power electrical systems such as headlights, air conditioning, or sound systems. It can also be used to assist the engine during acceleration, reducing the workload and boosting fuel economy.
There are three types of regenerative braking systems. They are:KERS (Kinetic Energy Recovery System)Hydraulic Hybrid SystemFlywheel-based Regenerative Braking System. The hydraulic hybrid system is the type of regenerative braking system that uses an electrohydraulic system.
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How does the compressive strength, impact resistance and plastic shrinkage resistance of concretes are effected by increased volüme % of fibers?
What advantages can we have from macro-polymer fibers in terms of earthquake resistant structures?
Compressive strength, impact resistance, and plastic shrinkage resistance are the most important properties of concrete. These properties are important for the durability of concrete structures. The properties of concrete can be improved by adding fibers to the mix.
Increasing the volume % of fibers can significantly improve the compressive strength, impact resistance, and plastic shrinkage resistance of concretes. The increase in volume % of fibers can result in an increase in the compressive strength of up to 30%. The impact resistance of concrete can also be improved by increasing the volume % of fibers.
The addition of fibers can increase the energy absorption of concrete and reduce the severity of impact loads. Moreover, the plastic shrinkage resistance of concrete can also be improved by increasing the volume % of fibers. This can prevent the formation of cracks due to plastic shrinkage and reduce the risk of early age cracking.
Macro-polymer fibers are useful for earthquake-resistant structures. They can help reduce the damage caused by earthquakes. The addition of macro-polymer fibers can increase the ductility of concrete, which allows the structure to deform without collapsing during an earthquake.
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6. a. Take care of any injured parties b. Secure the site c. investigates the cause. d. All of the above 7. 8 Sufficient working space of A step in an accident investigation includes: M.C. An employer must report ALL company accidents resulting in a death to OSHA. T/F ft needs to be maintained around electrical equipment. FILL IN
Step in an accident investigation: Accident investigations are important to determine the cause of an accident. It can help to determine why and how the accident occurred.
A step in an accident investigation includes taking care of any injured parties, securing the site, investigating the cause, and reporting the accident. All of the above are steps that should be taken in an accident investigation. Sufficient working space: An adequate working space needs to be maintained around electrical equipment.
This helps to prevent electrical hazards and can help to prevent accidents. Employer's reporting of company accidents:An employer must report ALL company accidents resulting in a death to OSHA. It is the responsibility of the employer to provide a safe working environment for their employees. OSHA is responsible for ensuring that employers provide a safe and healthy workplace for their employees. OSHA has regulations and standards that employers must follow to ensure the safety of their employees.
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Design a tied column with DL = 500 KN, LL = 650 KN. fc' = 20.7 MPa, fy = 276 MPa. Use pg= 0.03, 25 mm Ø main bars, and 10 mm Ø ties.
Tied columns are used when there is a lack of space, limited headroom, or heavy lateral loads on the structure. They can be designed to withstand any amount of compressive stress, and the ties help keep them from buckling.
Given DL = 500 KN, LL = 650 KN, fc' = 20.7 MPa, fy = 276 MPa, pg = 0.03, 25 mm Ø main bars, and 10 mm Ø ties, here's how to design a tied column:
Step 1: Find out the factored load. The total factored load on the column is calculated as follows: Pu = 1.2DL + 1.5LLPu = 1.2 × 500 + 1.5 × 650Pu = 1460 KN
Step 2: The radius of gyration, r, is determined by: r = (0.5 × diameter)/square root of area We have to take the larger value for L in this case since it is greater than the smaller value. Therefore: Effective length, Le = 3 meters Radius of gyration, r = 7.07 cmK = (3m / 7.07cm)K = 42.43
Step 3: Calculate the effective length. The effective length of a column is found using the following formula :L = (K * r)The effective length is found by substituting the values into the equation: L = 42.43 * 7.07L = 300 cm.
Step 4: Determine the minimum and maximum steel ratios. The minimum steel ratio for a tied column is 0.008. The maximum steel ratio is 0.4.
Step 5: Check for minimum steel percentage. The minimum steel percentage is calculated using the following formula:ρmin = 0.008 * (fc' / fy)ρmin = 0.008 * (20.7 / 276)ρmin = 0.0006The calculated steel percentage of 0.01 is greater than the minimum requirement, thus, it is okay. So the column design is okay.
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1.Usually an asphalt with 20mm aggregate is thicker than the one with only 10mm aggregate. a. False b. True 2. Stone mastic asphalt is different from fine gap graded asphalt because it has more coarse aggregate. a. False b. True 3. In decreasing relative noise levels of new bituminous surfacings, which one of the following is correct? a. Spray seal (7mm), fine gap graded asphalt, stone mastic asphalt b. Open graded asphalt, Stone asphalt, fine gap graded asphalt c. Fine gap graded asphalt, open graded asphalt, stone mastic asphalt d. Spray seals (10mm), stone mastic asphalt, dense graded asphalt 4 .One of the advantages of Dense garded asphalt is that most contractors and producers are familiar with its production and placement. a. False b. True 5.One of the advantages of Dense garded asphalt is that most contractors and producers are familiar with its production and placement. a. False b. True 6. Why is Stone mastic asphalt an expensive mix? a. It uses more binder b. It uses a lot of large aggregate C. It has more filler d. It has high resistance to shear force 7. Which one the following asphalt is best for good texture and high resistance to shear force? a. Dense graded asphalt b. Stone mastic asphalt c. Gap graded asphalt . d. Open graded asphalt 8. Which one the following is true about Dense Graded asphalt? a. It can accommodate high binder contents without becoming unstable and susceptible to rutting b. It has a low design air void content of about 3 to 7% C. It cannot be used for heavily trafficked road because it has a low shear resistance d. It has about 8 – 12 % of filler
1. False. Usually, asphalt with 10mm aggregate is thicker than the one with only 20mm aggregate.2. True. Stone mastic asphalt is different from fine gap graded asphalt because it has more coarse aggregate.3. c. Fine gap graded asphalt, open graded asphalt, stone mastic asphalt.
In decreasing relative noise levels of new bituminous surfacings, this is the correct order.4. True. One of the advantages of Dense graded asphalt is that most contractors and producers are familiar with its production and placement.5.
True. One of the advantages of Dense graded asphalt is that most contractors and producers are familiar with its production and placement.6. a. It uses more binder. Stone mastic asphalt is an expensive mix as it uses more binder.
7. b. Stone mastic asphalt is the best for good texture and high resistance to shear force.
8. a. It can accommodate high binder contents without becoming unstable and susceptible to rutting. Dense graded asphalt can accommodate high binder contents without becoming unstable and susceptible to rutting.
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Between the Standard Form of Building Contract and New Engineering Contract,
(a)Outline the differences in project controls on -
i. Time;
ii. Cost;
iii. Quality; and
iv. Parties' interests.
(b) Discuss the principal differences in approaches on contract control such as substantive and procedural entitlements.
The Standard Form of Building Contract (SBC) and the New Engineering Contract (NEC) are the two most widely used building contracts in the UK.
Time: In the SBC, project controls in terms of time may be tracked using progress reports or programmed, and time is often referred to as the contract period.
Cost: In the SBC, project controls in terms of cost are often administered via periodic payments made to the contractor based on a pre-determined schedule of payments or stage payments.
Quality: In terms of quality, the SBC necessitates compliance with specifications and tolerances specified in the contract, whereas the NEC emphasizes the contractor's obligation to provide work of the requisite standard to accomplish the works' outcomes.
Parties' interests: The SBC's project controls are often weighted in the employer's favor, with little power given to the contractor, while the NEC provides greater flexibility in decision-making and negotiation between the parties.
Substantive Entitlements: These are the rights given to the parties under the contract and will be related to a party's entitlement to payment or the ability to terminate the contract. This is to ensure that both parties receive a fair deal.
Procedural Entitlements: These are procedures that need to be followed in the event of a dispute or disagreement over the contract's substantive terms.
SBC and NEC have their differences in terms of project control, and they are both commonly used building contracts in the UK.
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A construction project manager is using PERT method to determine the expected project completion time for the construction of a new hospital. The expected project completion time is found to be 21 months and project variance is 4 months. a) What is the probability that the project will be completed within 17 months? b) What is the probability that the project will be completed within 23 months? c) What is the due date that yields a 95% chance of completion?
PERT (Program Evaluation Review Technique) is a statistical project management tool utilized to measure time taken to complete tasks in the development of a new product.
In this method, the optimistic time, the most probable time, and the pessimistic time are all taken into account to calculate expected time. Additionally, the time variability of the project is calculated.Using the PERT method to determine the expected project completion time for the construction of a new hospital, we are given that the expected project completion time is 21 months and the project variance is 4 months. A) The probability that the project will be completed within 17 months is required. Using the formula
, z = (X - μ) / σ, where X = 17, μ = 21, and σ = sqrt(4) = 2z = (17-21) / 2 = -2P(z) = 0.0228The probability that the project will be completed within 17 months is 0.0228.b) We are required to find the probability that the project will be completed within 23 months. Using the formula, z = (X - μ) / σ, where X = 23, μ = 21, and σ = sqrt(4) = 2z = (23-21) / 2 = 1P(z) = 0.8413The probability that the project will be completed within 23 months is 0.8413c) We are required to find the due date that yields a 95% chance of completion. Using the formula, z = (X - μ) / σ, we can find the z-score that corresponds to a 95% chance of completion. z = 1.645The equation that relates due date, mean completion time, and standard deviation is X = μ + zσX = 21 + 1.645(2)X = 24.29The due date that yields a 95% chance of completion is 24.29 months.
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29. A construction company has grown income of 8 million dollars. All expenses except capital expenditure is 4.2 million. Depreciation allowance is $100,000. What is taxable income?
a. $4.3 million
b. $3.4 million
c. $3.7 million
d. $7.3 million
30. Use Question 29. What is the amount of tax?
a. $1.258 million
b. $2.158 million
c. $5.128 million
d. $8.125 million
29. Taxable Income is the income on which tax is paid by an individual or a corporation. Taxable income is calculated by subtracting allowable deductions from gross income. Therefore, from the given data, taxable income can be calculated as:
Gross Income = $8 million All expenses except capital expenditure = $4.2 million Depreciation allowance = $100,000Therefore, taxable income = $8 million - ($4.2 million + $100,000)= $8 million - $4.3 million = $3.7 million Hence, the taxable income is $3.7 million. Option (c) is correct.30.
Tax is an amount of money that people have to pay to the government, usually based on a percentage of their income. The amount of tax paid depends on the tax rate and the taxable income. Therefore, from the given data, the amount of tax can be calculated as:
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A tractor for over-the-road hauling is purchased for $90,000.00. It is expected to be of use to the company for 6 years, after which it will be salvaged for $3,600.00. Calculate the depreciation deduction and the unrecovered investment during each year of the tractors life.
a. Use straight-line depreciation. Provide depreciation and book value for year 6.
Depreciation for year 6 = $_________________
book value for year 6 = $________________
b. Use declining-balance depreciation, with a rate that ensures the book value equals the salvage value. Provide depreciation and book value for year 6.
Depreciation for year 6 = $________________
book value for year 6 = $___________________
c. Use double declining balance depreciation. Provide depreciation and book value for year 6.
Depreciation for year 6 = $________________
book value for year 6 = $_________________
d. Use double declining balance, switching to straight-line depreciation. Provide depreciation and book value for year 6.
Depreciation for year 6 = $_________________
book value for year 6 = $___________________
a) Depreciation and Book Value for Year 6 (Straight-line Depreciation):
[tex]($90,000 - $3,600) / 6Depreciation = $14,900[/tex]
Book Value for Year 6 = [tex]$90,000 - ($14,900 × 5)[/tex]
Book Value for Year 6 = $11,500
b) Depreciation and Book Value for Year 6 (Declining-balance Depreciation):Depreciation Rate = 2 / Useful Life= [tex]($90,000 - $3,600) x 33.33%[/tex]Depreciation = $28,787.04
Book Value for Year 6 = $3,600
c) Depreciation and Book Value for Year 6 (Double Declining-balance Depreciation):Depreciation Rate = 2 x Straight-line Depreciation = [tex]($90,000 - $3,600) x 33.33%[/tex]Depreciation = $28,787.04
Book Value = Cost - Accumulated Depreciation Book Value for Year 6 = $3,600
d) Depreciation and Book Value for Year 6 (Double Declining-balance Depreciation, Switching to Straight-line Depreciation):
Depreciation Rate = 2 x Straight-line Rate Depreciation Rate = 2 x 1/6Depreciation Rate = 33.33%Depreciation for Year 1 = Book Value x Depreciation Rate
Depreciation for Year 1 = [tex]($90,000 - $3,600) x 33.33%[/tex]Depreciation for Year 1 = $28,787.0
Depreciation for Year 6 =[tex]$11,500 x 20%[/tex]Depreciation for Year 6 = $2,300
Book Value for Year 6 = [tex]$90,000 - ($28,787.04 + $2,300)[/tex]Book Value for Year 6 = $58,912.96
The required answers for the above-mentioned question are: Depreciation for year 6 using straight-line depreciation = $2,300Book value for year 6 using straight-line depreciation = $58,912.96
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