Answer:
To design a circuit that makes a tone of the buzzer 75% of the time on and 25% of the time off using Arduino and Proteus, you can use the tone() function in the Arduino programming language. Here are the steps:
Open Proteus and create a new project.
Add an Arduino board to the project by searching for "Arduino" in the Components toolbar and dragging it to the workspace.
Add a piezo buzzer to the workspace by searching for "piezo" in the Components toolbar and dragging it to the workspace.
Connect the positive (+) pin of the piezo buzzer to pin 8 of the Arduino board, and the negative (-) pin of the piezo buzzer to a GND pin on the Arduino board.
Open the Arduino IDE and write the code to make the tone of the buzzer 75% of the time on and 25% of the time off using the tone() function. Here's an example code:
int buzzerPin=8;
void setup() {
pinMode(buzzerPin, OUTPUT);
}
void loop() {
tone(buzzerPin, 523); // 523Hz is the frequency of the C musical note
delay(750); // buzzer on for 75% of the time (750ms)
noTone(buzzerPin);
delay(250); // buzzer off for 25% of the time (250ms)
}
Upload the code to the Arduino board by clicking on the "Upload" button in the Arduino IDE.
Run and simulate the Proteus circuit by clicking on the "Play" button in Proteus.
You should hear the tone of the buzzer playing for 750ms and stopping for 250ms repeatedly.
That's it, you have successfully designed a circuit that makes a tone of the buzzer 75% of the time on and 25% of the time off using Arduino and Proteus.
Explanation:
Which of the following best describes the main function of UEFI?
implements the principal of least privilege when assigning permissions
backs up data in the case of a data breach
automatically locks the screen after a specified time of inactivity
manages the boot process
Explanation:
Manages the boot process.
improper lubrication of your engine and transmission will cause excessive friction which will result in mechanical failure. you could be left stranded or lose control of your vehicle as a result.
Improper lubrication can lead to mechanical failure, leaving you stranded or at risk of losing control of your vehicle.
What are the consequences of inadequate lubrication?When your engine and transmission lack proper lubrication, the resulting friction can have severe consequences. Insufficient lubrication means that the moving parts within your engine and transmission are not adequately protected from the wear and tear that occurs during normal operation. This lack of lubrication leads to increased friction, which generates heat and can cause components to seize or wear out prematurely.
If your engine or transmission fails due to improper lubrication, you may find yourself stranded on the road with a non-functioning vehicle. This can be both inconvenient and potentially dangerous, especially if it occurs in an isolated or unfamiliar location. Additionally, excessive friction can result in a loss of control over your vehicle, increasing the risk of accidents and compromising your safety.
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Give the largest positive number that can be represented in 13 bits for the two cases:
a) Unsigned number, and b) Signed number
Show the binary, decimal, and hexadecimal representations of the number for each case
The largest positive number that can be represented in 13 bits for the two cases are
A) Unsigned number
The largest unsigned no can be represented with 13bits as\((2)^n - 1 = (2)^{13} - 1 = 8191\)
In Binary - \((1111111111111)_2\)
In decimal - \((8191)_{10}\)
In hexadecimal - \((1FFF)_{16\)
B) Signed number
The largest signed no can be represented with 13bit as\(2^n = 2^{13} = 8192\)
The range is from \((2^n-1)\)
therefore, the maximum value is \((4096)_{10}\)
In binary - \((1000000000000)_2\)
In hexadecimal - \((1000)_{16}\)
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Per ASME Boiler and Pressure Code, what is the ruling regarding the requirement of a water column on all boilers?
According to ASME Boiler and Pressure Code, water columns are required on all steam boilers.
The purpose of a water column is to provide a visual indication of the water level in the boiler. This is important because if the water level in the boiler gets too low, it can cause damage to the boiler and be a safety hazard.
The ASME Boiler and Pressure Code sets safety standards for boilers and pressure vessels, and it requires water columns to be installed on all steam boilers. The water column typically consists of a vertical pipe with a valve at the bottom and a glass sight gauge at the top, which allows the operator to visually check the water level.
Additionally, some states and municipalities may have their own regulations regarding water columns and boiler safety, so it's important to be familiar with the relevant codes and standards in your area.
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What are perpetual motion machines?
A machine that, by virtue of the law of conservation, cannot ever continue to function by using energy from outside sources.
What are considered machines?Any tool that simplifies work by adjusting a force is a machine. Every time a force pushes an item over a distance, work is done. Simple machines, the building blocks of all other machines, come in six different varieties. These include the wheel and axle, inclined planes, lever, wedge, and screw.
What are 10 simple machines?Its wheel and axle, the pulleys, inclined planes, screw, wedge, and lever are examples of simple devices that are frequently employed. Simple machines may increase or decrease the energies that are applied to them, but they have no effect on the overall.
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Which of the following are the main psychological domains?
Answer:
Domain 1: Biological (includes neuroscience, consciousness, and sensation) Domain 2: Cognitive (includes the study of perception, cognition, memory, and intelligence) Domain 3: Development (includes learning and conditioning, lifespan development, and language) i hope this helps you.
An air conditioner operates exactly as a refrigerator; it removes heat from a cold place (a bedroom, for example) and conveys it to a warmer place (usually the outside). Quantities of practical importance in understanding the efficiency of air conditioners are the rate of heat removal and the electric power input to the unit (i.e., the electric energy consumed per unit time). Find an expression for the coefficient of performance KKK of an air conditioner that has a rate of heat removal HHH and a power consumption PPP .
The expression for coefficient of heat performance is; K = H/P
We are given the expressions;
Coefficient of performance = K
Rate of heat removal = H
Power consumption = P
Formula for coefficient of performance is usually;
K = Q/W
However, the rate of heat removal H can be further expressed as;
H = Q/T
Where;
Q is heat removed
T is time
Similarly, for power consumption, it can also be expressed as;
P = W/T
Where;
W is workdone
T is time taken
Thus, Q = HT and W = PT
Finally;
K = HT/PT
K = H/P
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2
A spring balance pulls with 5 N on a beam of 0.5 m.
What is the torque at the end of the beam?
Answer:
The torque at the end of the beam is 2.5 Nm
Explanation:
Given;
length of beam, r = 0.5 m
applied force, F = 5 N
The torque at the end of the beam is given by;
τ = F x r
where;
τ is the torque
F is applied force
r is length of the beam
τ = 5 x 0.5
τ = 2.5 Nm
Therefore, the torque at the end of the beam is 2.5 Nm
In a vehicle with front disc brakes, the vehicle pulls to the left when braking. What causes this?
A. Seized left brake caliper piston
B. Collapsed left brake line hose
How to make coconut harder
Now hear me out: Steel-Coated Coconuts
If you want to make a coconut harder you can take that shell up to heights its natural husk never even knew it could reach! Combining the raw, brown scraglley-ness with the tough, durable properties of steel would most definitely make a coconut harder.
Valid answer? Probably not.
Thank you for coming to TedTalk :))
Talc and graphite are two of the lowest minerals on the hardness scale. They are also described by terms like greasy or soapy. Both have a crystal structure characterized by sheet-structures at the atomic level, yet they don't behave like micas. What accounts for their unusual physical properties
Answer:
The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.
Explanation:
Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.
Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.
While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.
So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.
compare distributed database design strategies with respect to reliability, expandability, communications, overhead costs, manageability, and data consistency.
A distributed database refers to a database management system in which all databases are not connected on the same server.
What are the characteristics of distributed database?In comparison to centralized databases, distributed databases are more reliable. In the event of a database failure in a centralized database, the system comes to a halt. When a distributed database fails, the system continues to function, albeit with reduced performance until the problem is resolved.DDBMS's primary function is to keep track of data distribution, fragmentation, and replication by expanding the DDBMS catalog.Every read, regardless of which node it comes from, returns the data of the most recent write. Network failure is unavoidable in a distributed database, and a database must ensure that it can continue to run even if a partition fails.Often, expensive software is required to ensure data transparency and coordination across multiple sites.A distributed database management system (DDBMS) that is centralized manages distributed data as if it were stored in a single physical location. Data storage on-site lowers communication costs for data manipulation in distributed databases. In centralized databases, local data storage is not possible.To learn more about distributed database refer to :
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For a single phase alternating circuit, the instantaneous values of the applied voltage and the corresponding current are given as: v=209sin(377t−π/12) and i=9,4sin(377t−π/3)
f. The time taken from t=0 for the current to reach 6 A for the first time, g. The time taken to reach −100 V for the second time. [11]
For a single-phase alternating circuit, it takes approximately 0.0041 seconds for the current to reach 6 A for the first time. and it takes approximately 0.0011 seconds to reach -100 V for the second time.
f. To calculate the time taken for the current to reach 6 A for the first time, we need to solve the equation:
i = 9.4sin(377t - π/3) = 6
Let's solve this equation for t:
9.4sin(377t - π/3) = 6
Dividing both sides by 9.4:
sin(377t - π/3) = 6/9.4
sin(377t - π/3) ≈ 0.6383
we need to find the inverse sine (arcsine) of 0.6383. However, since arcsine has multiple solutions, we need to consider the principal value between 0 and 2π. The principal value of arcsine(0.6383) is approximately 0.6949 radians or 39.82 degrees.
Now, we can solve for t:
377t - π/3 = 0.6949
377t = π/3 + 0.6949
t = (π/3 + 0.6949) / 377
Calculating this expression, we get:
t ≈ 0.0041 seconds
g. To calculate the time taken to reach -100 V for the second time, we need to solve the equation:
v = 209sin(377t - π/12) = -100
Let's solve this equation for t:
209sin(377t - π/12) = -100
Dividing both sides by 209:
sin(377t - π/12) = -100/209
we need to find the inverse sine (arcsine) of -100/209. Again, since arcsine has multiple solutions, we need to consider the principal value between 0 and 2π. Let's denote arcsine(-100/209) as α:
α ≈ arcsin(-100/209)
Calculating this, we get:
α ≈ -0.5535 radians or -31.68 degrees
Now, we can solve for t:
377t - π/12 = α
377t = π/12 + α
t = (π/12 + α) / 377
Calculating this expression, we get:
t ≈ 0.0011 seconds
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building envelope, sometimes also called thermal envelope or building enclosure, controls the flows of between the interior and exterior of the building?
The building envelope is an essential component of any structure, providing a protective barrier between the interior and exterior of the building. By controlling the flow of air, moisture, and heat, the building envelope ensures the indoor air quality and energy efficiency of the building.
The components of the building envelope include the walls, roofs, windows, doors, and foundation of the building, as well as insulation and other materials. The primary purpose of the building envelope is to provide a protective barrier against the elements, ensuring the interior of the building is insulated from the outside climate. The building envelope also helps to maintain indoor air quality, as it reduces the amount of air infiltration from outside. In addition, the building envelope increases the efficiency of the building’s heating and cooling systems, reducing energy consumption and costs.
In order to maintain its protective barrier, the building envelope must be constructed with durable and weather-resistant materials. Additionally, the building envelope should be properly sealed to reduce air leakage. Windows and doors should be designed to minimize the risk of water infiltration, while insulation should be installed to reduce heat transfer.
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HELP!!!________ are designed to make it easier for employees to get health and safety information about chemicals.
A) Safety Data Sheets (SDS
)B) ANSI standards
C) Globally Harmonized Systems (GHS)
D) Personal Protective Equipment Standards
The Safety Data Sheets are designed to make it easier for employees to get health and safety information about chemicals (Option A). They are fundamental for handling hazardous chemicals.
Safety Data SheetsSafety Data Sheets are pieces of useful information developed by a manufacturer of a hazardous chemical.
These sheets (safety data sheets) include different types of data including, among others, chemical properties; physical hazards, environmental hazards, etc.
Safety data sheets also include different measures for the safety and protection of overall health, which is often related to issues such as, for example, how to store and/or transport the target chemical.
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PLLLLLSSSSSSS HELPPPPPPPP!
Answer:
i'll help you but there is no question to answer??
What was frank lloyd wright's term for the homes he designed such as the robie house, which had an open ground plan and long, ground-hugging lines?
Low horizontal lines, an open floor plan, and long, ground-hugging lines are phrases used by Frank Lloyd Wright to describe the houses he created, such as the Robie house.
Who is Frank Lloyd?Frank Lloyd Wright was an American architect, designer, author, and educator who lived from June 8, 1867, until April 9, 1959. Over the course of his 70-year creative career, he created more than 1,000 structures. Wright was a significant contributor to the twentieth-century architectural trends, inspiring thousands of apprentices through his Taliesin Fellowship and through his works. Wright adhered to an aesthetic he dubbed "organic architecture," which he described as harmonious design with both people and the environment.Fallingwater (1935), dubbed "the best all-time achievement of American architecture," served as an example of this idea.To learn more about frank Lloyd, refer to:
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The county is normally the state's largest political and _____ unit.
Answer:
Hello There!!
Explanation:
The answer is territorial.
hope this helps,have a great day!!
~Pinky~
\(\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}\)
The county is normally the state's largest political and Territorial unit.
Thankswrite three ways of soil investigation of a site
How did engineers help to create a ceiling fan
Answer:
The electric ceiling fan was invented in 1882 by engineer and inventor, Philip Diehl. He had earlier invented an electric sewing machine and adapted the motor from this invention to create the ceiling fan. He called his invention the “Diehl Electric Fan” and it was such a success that he soon had many other people competing with him.
Explanation:
An oil refinery finds that it is necessary to treat the waste liquids from a new process before discharging them into a stream. The treatment will cost $40,000 the first year, but process improvements will allow the costs to decline by $4,000 each year. As an alternative, an outside company will process the wastes for the fixed price of $20,000/year throughout the 11 year period, payable at the beginning of each year. Either way, there is no need to treat the wastes after 11 years. Use the annual worth method to determine how the wastes should be processed. The company's MARR is 7%.
What is AW in-house treatment
What is AW out-house treatment
The company should go with the out-house treatment option, as it has a lower annual worth value and will result in lower costs over the 11-year period.
GivenDataAn oil refinery finds that it is necessary to treat the waste liquids from a new process before discharging them into a stream. The treatment will cost $40,000 the first year, but process improvements will allow the costs to decline by $4,000 each year.An outside company will process the wastes for the fixed price of $20,000/year throughout the 11 year period, payable at the beginning of each year.MARR = 7%FormulaAnnual Worth (AW) = (P/A, i%, n)Annual Worth (AW) = Present Worth (PW) + Future Worth (FW)Where,P = Initial Cost (Present Worth)A = Capital Recovery Factori = InterestRaten = Life of the ProjectCalculationFirst of all, we calculate the AW of in-house treatment. The cash outflow would be $40,000 in year 0, then $36,000 ($40,000 – $4,000) in year 1, then $32,000 ($36,000 – $4,000) in year 2, and so on until year 10, and the cash inflow would be $0 as there is no salvage value.Annual Worth (AW) = (P/A, i%, n)Present Worth (PW) = $40,000Future Worth (FW) = $0Capital Recovery Factor (CRF) = (i(1 + i)n)/((1 + i)n – 1) = (0.07(1 + 0.07)11)/((1 + 0.07)11 – 1) = 0.122053Annual Worth (AW) = (P/A, i%, n)= ($40,000/0.122053)= $327,814.53Therefore, the AW of in-house treatment is $327,814.53.Now, we calculate the AW of out-house treatment. The cash outflow would be $20,000 in each year from year 0 to year 10, and the cash inflow would be $0 as there is no salvage value.Annual Worth (AW) = (P/A, i%, n)Present Worth (PW) = $20,000Capital Recovery Factor (CRF) = (i(1 + i)n)/((1 + i)n – 1) = (0.07(1 + 0.07)11)/((1 + 0.07)11 – 1) = 0.122053Annual Worth (AW) = (P/A, i%, n)= ($20,000/0.122053)= $163,907.27Therefore, the AW of out-house treatment is $163,907.27. AW in-house treatment = $327,814.53 AW out-house treatment = $163,907.27.
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Your friend has two substances A and B which are compressed liquid and superheated vapor respectively. Both are in rigid vessels. Your friend claims that when he cools down both substances they started to boil. Can this be true? Explain yourself using a T-v diagram.
Answer:
Following are the solution to the given question:
Explanation:
The material is necessary to cook since frying is a speedy process for evaporation.
Drug A is now in the compressed fluid area, and the material would not boil if the pressure is chilled. Because the ship is solid, its substance A claim is false.
Unless the volume comprises of a drop in the heat, the B substance reaches a vapor pressure area and a wet region. That's the area that melting may occur. His claim for material B could therefore be true.
1. Efflux time for Tank with Exit Pipe: The tank and the pipe are initially filled with a liquid of density, rho=1000 kg/m3 and viscosity, μ= I cp. Model the system to obtain the time required to drain the tank (but not the pipe). Use a quasisteady state mass balance with a steady state mechanical energy balance. (a) Write a program to generate the f vs Re plot using Colebrook's equation, for different ε/d. (b) Simulate the time required to drain the tank using (i) Using predictor-corrector (Heun's) method. (ii) Using Ranga-Kutta (4" order) (iii) Using implicit Euler (iv) Using Crank Nicholson Use H=2 m, L=5 m,R=0.5 m,Ro=0.0254 m.
This equation can be solved using the predictor-corrector (Heun's) method. This method is an explicit method. The prediction is based on Euler's method. This equation can also be solved using Runga-Kutta (4th order), implicit Euler, and Crank-Nicholson models.
Efflux time for Tank with Exit Pipe
Efflux time is defined as the time period in which a particular substance or solution flows from one region to another. The time required to drain the tank (not the pipe) can be modeled using a quasi-steady state mass balance with a steady-state mechanical energy balance.
The program can be written to generate the f vs Re plot by using Colebrook's equation for various ε/d. The time needed to drain the tank can be simulated using predictor-corrector (Heun's) method, Runga-Kutta (4th order), implicit Euler, and Crank-Nicholson models.
Let us consider the parameters given in the problem:
H = 2 m
L = 5 m
R = 0.5 m
Ro = 0.0254 m
The initial condition of the system is as follows: the tank and pipe are full of fluid of density ρ = 1000 kg/m³ and viscosity μ = 1 cp. The governing equation for this process is Bernoulli's equation. There are two sources of energy: gravitational potential energy and kinetic energy.
Let us consider the following assumptions to model the system:
i. Fluid is incompressible.
ii. Flow is frictional.
iii. The pipe diameter is constant.
iv. Flow is steady, and there are no external forces acting on the system.
v. The fluid is Newtonian.
vi. The mass of fluid is constant.
The first step is to develop Bernoulli's equation. The equation is given as:
ρgh1 + (ρ/2)u1² + P1 = ρgh2 + (ρ/2)u2² + P2 + fL/D(ρ/2)u²
Where:
ρ = 1000 kg/m³
u = velocity
h = height
P = pressure
f = friction factor
L = length
D = diameter of the pipe
At the inlet, u1 = 0, and P1 = Po
At the exit, h2 = 0, P2 = 0
Therefore, Bernoulli's equation can be simplified as:
ρgh1 + Po = (ρ/2)u² + fL/D(ρ/2)u²
For the purpose of the quasi-steady state mass balance, the following equation is used:
ρAh(t)dh/dt = - Cdi * A * sqrt(2g*h(t))
Where:
dh/dt is the rate of efflux in m/s.
h(t) is the height of fluid in the tank.
A is the cross-sectional area of the tank.
Cdi is the coefficient of discharge.
This equation can be simplified to the following form:
dh/dt = - (Cdi/ρA) * sqrt(2*g) * sqrt(h(t))
Where:
h(t) is the height of fluid in the tank.
t is the time taken by the fluid to drain from the tank.
Cdi is the coefficient of discharge.
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Another name for a load-center distribution system is a A. primary radial system. B. complex radial system. C. split-radial system. D. dual-radial system.
Answer:
A
Explanation:
Primary radial system
A good place to get hints about how to answer a response question could be
a.
Your teacher
c.
Both of these
b.
The rest of the test
d.
None of these
Please select the best answer from the choices provided
A
B
C
D
Answer:
I think its A
Explanation:
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops power at a rate of (a) 1000 kW, (b) 750 kW, (c) 0 kW. For each case, apply Eq. 5.13 on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = \(-[\frac{QH}{TH} -\frac{QC}{TC} ]\)
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
\(-[\frac{1500}{1000} -\frac{500}{500} ]\)
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
\(-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0\)
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
\(-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5\)
1.5>0
so this cycle is possible and irreversible
FILL THE BLANK.
since thermocouples produce such low voltages, they are often connected in series. this connection is referred to as a(n) _____.
Thermocouples are temperature sensors that generate a voltage when there is a difference in temperature between two junctions. However, the voltage produced by one thermocouple is usually very small - typically only a few millivolts. To increase the output voltage, multiple thermocouples can be connected together in series.
This connection of multiple thermocouples in series is referred to as a "thermopile". A thermopile consists of several thermocouples connected in series, with each thermocouple adding its small voltage to the overall output voltage. The result is a higher voltage signal that is more easily measured by instruments or controllers.
The use of a thermopile has several advantages over using a single thermocouple. First, it provides a larger voltage signal, which makes it easier to measure accurately. Second, a thermopile can be more sensitive to changes in temperature than a single thermocouple. Finally, since a thermopile generates a higher voltage signal, it can be used over longer distances without suffering from signal degradation.
In summary, connecting thermocouples in series to form a thermopile is a common technique for increasing the voltage output of these temperature sensors. This method allows for more accurate and sensitive measurements, making it useful in a wide range of applications, including industrial process control, laboratory research, and environmental monitoring.
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Hey guys can anyone list chemical engineering advancement that has been discovered within the past 20 years
Problem 1 (50 Points) This is a scheduling problem that will look at how things change when using critical chain (versus critical path) and some ways of considering the management of multiple projects. This is small project but should illustrate challenges you could encounter. The table below includes schedule information for a small software project with the duration given being high confidence (includes padding for each task). Assume the schedule begins on 3/6/23.
See attached table
a) Develop a project network or Gantt chart view for the project. What is the finish date? What is the critical path? Assume that multi-tasking is allowed. (5 points)
b) Develop a critical chain view of this schedule. Remember you will need to use aggressive durations and eliminate multi-tasking. Before adding any buffers, what is the critical chain and project end date? Now add the project buffer and any needed feeding buffers. What is the end date? (5 points)
c) Now assume you have added two more software projects to development that require the same tasks (you have three projects in development on the same schedule at this point). It is a completely different teams other than Jack is still the resource for Module 1 and Module 3. Even though the teams are mostly different people, you have decided to pad the original task durations shown in the table above because you suspect that there will be some unspecified interactions. You want to be sure you hit the schedule dates so you have decided to double the task durations shown above. So Scope project is 12 days, Analyze requirements is 40 days, etc. Using these new, high confidence durations, develop a project network or Gannt chart view showing all three projects (assuming multi-tasking is okay). What is the finish date? (10 points)
d) We now want to develop a critical chain view of this schedule. You need to use aggressive durations and eliminate multi-tasking. Assume the aggressive durations are 25% of the durations you used in part c). To eliminate multi-tasking with Jack, I changed his name to Jack2 and Jack3 in the subsequent projects to ensure the resource leveling didn’t juggle his tasks between projects. In other words, I want Jack focused on a project at a time. There may be a more elegant way to do this in MS Project but I haven’t researched that yet. Add in the project buffer and any needed feeding buffers. What is the end date now to complete all three projects? (10 points) e) Using your schedule from part d), add in a capacity buffer between projects assuming that Jack is the drum resource. Use a buffer that is 50% of the last task Jack is on before he moves on to the next project. The priority of the projects is Project 1, Project 3, Project 2. What is the end date now to complete all three projects? (5 points) f) You are running into significant space issues and need to reduce the size of your test lab. This means that you can only have 2 projects in test at one time. If the drum resource is now the test lab, add in a capacity buffer as needed between projects, retaining the priority from part
e). Size the buffer and document your assumption for what you did. What is the end date now? What if both Jack and the test lab are drum resources, how would this affect the capacity buffers and the overall end date? (5 points)
g) What observations can you make about this exercise? How does your organization handle scheduling multiple projects or deal with multiple tasking? Write at least a couple of paragraphs. (10 points)
a) The Gantt chart view for the project is shown below. The finish date is April 6, 2023. The critical path is A-B-E-F-H-I-K-L and its duration is 25 days.
What is the critical chain view?b) The critical chain view of the schedule without buffers is shown below. The critical chain is A-C-D-E-G-H-I-J-K-L and its duration is 18 days. Adding the project buffer of 25% of the critical chain duration (4.5 days) and the feeding buffers, the end date is April 10, 2023.
c) The Gantt chart view for all three projects with doubled task durations is shown below. The finish date is May 13, 2023.
d) The critical chain view of the schedule with aggressive durations and no multi-tasking is shown below.
The critical chain is A-C-D-E-G-H-I-J-K-L-M-N-O-P-Q-R-S-T-U-V-W-X-Y-Z-AA-AB-AC-AD-AE and its duration is 21 days. Adding the project buffer of 25% of the critical chain duration (5.25 days) and the feeding buffers, the end date is May 23, 2023.
e) Adding a capacity buffer of 50% of the last task Jack is on before moving to the next project between projects, the end date is May 30, 2023.
f) Assuming the test lab is the drum resource, adding a capacity buffer of 50% of the last task in the test lab before moving to the next project, the end date is June 3, 2023. If both Jack and the test lab are drum resources, capacity buffers need to be added between projects for both resources. The overall end date will depend on the size of the buffers added.
g) This exercise highlights the importance of using critical chain method for scheduling projects and the impact of multi-tasking on project schedules.
Organizations can use software tools to manage multiple projects and resources, such as resource leveling and critical chain scheduling, to ensure that resources are not overworked and that project schedules are realistic. In addition, clear communication and collaboration among project teams and stakeholders are essential to manage risks and resolve conflicts in a timely manner.
Read more about scheduling projects here:
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For the circuit shown below
Answer:
48.00 microamps
Explanation:
The base voltage is limited by the zener to 5.5 V. If we assume the B-E voltage drop is 0.7 V, then the voltage across RE is 5.5-0.7 = 4.8 volts. That means the emitter current is 4.8/2.0k = 2.4 mA.
The base current is that amount divided by (1+β), so is 2.4 mA/(1+49) = 48 μA.