Transcribed image text: Question 4 If a Haskell function £ have a type of f :: Int -> Int -> (Int, Int) Then the type of f 3 is Of 3 :: Int -> Int Of 3 :: Int -> (Int, Int) O £ 3 :: (Int) -> (Int, Int) Of 3 :: Int -> Int -> (Int) 1 pt Question 5 The following is the prototype of the printf function in C: int printf (char *format, ...); According to this prototype, the printf functions takes Oat least two (2) exactly one (1) exactly two (2) at least one (1) parameter(s). 1 pts Question 8 Given the following Horn clauses: X-A, B Y-X Which one can we obtain? OA, B Y OY A, B OY B OY A

Answer:

  R = 0

Explanation:

The output R is the result of the function ...

  R = (P+Q)·Q'

This simplifies to ...

  R = PQ' + 0

  R = PQ'

The result R will be the AND of P=0 and Q'=(1-1)=0. Any AND function with a 0 input will have a 0 output.

  R = 0

Highway safety is a multifaceted approach that involves a combination of engineering, enforcement, and education strategies. The design features contribute to the overall safety of expressways, but driver attentiveness, adherence to traffic laws, and responsible driving behavior are equally crucial for safe travel on these roadways.

Highway safety design features on expressways that help make them safer roadways to travel include:

1. Controlled Access: Expressways are designed with limited access points, such as ramps and interchanges, to regulate the flow of vehicles entering and exiting the roadway. This helps reduce the potential for collisions and congestion.

2. Median Barriers: Medians with physical barriers, such as concrete walls or cable barriers, are installed to separate opposing traffic. These barriers help prevent head-on collisions and minimize the severity of accidents.

3. Clear Zone Areas: Clear zones are areas adjacent to the roadway that are kept free from fixed objects, such as trees or poles. This provides a buffer zone and allows drivers to recover safely if they veer off the road.

4. High Visibility Markings: Expressways often have clear and highly visible lane markings, signage, and pavement markings. These aids help drivers navigate the roadway, change lanes safely, and understand upcoming exits and interchanges.

5. Lighting: Adequate lighting is installed along expressways to enhance visibility during nighttime conditions. Well-lit roadways improve driver awareness and reduce the risk of accidents.

6. Guardrails: Guardrails are installed along certain sections of the expressway, such as curves or areas with significant grade changes, to prevent vehicles from leaving the road and reduce the impact severity of collisions.

7. Emergency Pull-off Areas: Expressways may have designated emergency pull-off areas where drivers can safely stop in case of emergencies, breakdowns, or for accessing emergency services.

8. Traffic Control Devices: Intersections, interchanges, and merging areas on expressways are equipped with traffic control devices such as signals, yield signs, and merge signs to facilitate the safe movement of vehicles.

9. Advanced Warning Systems: Expressways may incorporate advanced warning systems, such as electronic message signs, variable speed limit signs, and intelligent transportation systems, to provide real-time information to drivers about hazards, congestion, or incidents ahead.

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Answer: C) Collective action problem

Explanation:

To find the largest principal stress at the given point, we need to analyze the Mohr's circle. Mohr's circle is a graphical method used to determine principal stresses and their orientations in a plane stress state.

From the given Mohr's circle, we can see that the largest principal stress occurs at the point where the circle intersects the x-axis. This point represents the maximum tensile stress.

To find the value of the largest principal stress, we need to read the corresponding value on the x-axis. Let's call this value σ1.

Therefore, the largest principal stress at this point is σ1.

Please note that without a visual representation of the Mohr's circle, it is not possible to provide a specific numerical value for σ1. However, by analyzing the circle, you can determine the largest principal stress based on its position relative to the x-axis.

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(a) The temperature of the exiting mixture is 300.8 8C.

(b) The rate of entropy production is 0.767 kW/K.

(c) The rate of exergy destruction is 177.5 kW.

Air at 50 8C, 1 atm and a volumetric flow rate of 60 m3/min mixes with helium entering as a separate stream at 120 8C, 1 atm and a volumetric flow rate of 25 m3/min. A single mixed stream exits at 1 atm. Ignoring kinetic and potential energy effects, determine for the control volume.

(a) the temperature of the exiting mixture, in 8C. (b) the rate of entropy production, in kW/K. (c) the rate of exergy destruction, in kW, for T0 5 295 K.The given problem can be solved by the application of the energy and mass balance equations as well as the second law of thermodynamics.

The rate of mass flow (m) of air (ṁ_air) and helium (ṁ_he) can be calculated using the given volumetric flow rate and the density of air and helium.The mass flow rate of air (mṁ_air) = Volumetric flow rate of air (V_air) × Density of air (ρ_air) = 60 m3/min × 1.2 kg/m3= 72 kg/min

The mass flow rate of helium (mṁ_He) = Volumetric flow rate of helium (V_He) × Density of helium (ρ_He) = 25 m3/min × 0.166 kg/m3 = 4.15 kg/min.The mass flow rate of the mixture is the sum of the mass flow rates of the air and helium.

Therefore, the mass flow rate of the mixture (mṁ_mixture) = mṁ_air + mṁ_He = 72 + 4.15 = 76.15 kg/min(a) Calculation of the temperature of the exiting mixture:By applying the energy balance equation for steady-state processes, the energy input equals the energy output. That is,Q_in = Q_outor, mṁ_air Cpa T1 + mṁ_He Cp_He T2 = mṁ_mixture Cp T3

where, Cpa and Cp_He are the specific heat capacities of air and helium at constant pressure respectively.Cp is the specific heat capacity of the mixture at constant pressure.T1 and T2 are the temperatures of air and helium before mixing.T3 is the temperature of the mixture after mixing.

Substituting the given values into the above equation and solving for T3, we get;72 × 1.005 × (508 − T3) + 4.15 × 5.193 × (1208 − T3) = 76.15 × 1.128 × (T3 − 298)T3 = 573.8 K = 300.8 8C

Therefore, the temperature of the exiting mixture is 300.8 8C.

(b) Calculation of the rate of entropy production:The rate of entropy production can be calculated using the second law of thermodynamics. The entropy generation (S_gen) is given by the equation:S_gen = ṁ_mixture × s_mixture − ṁ_air × s_air − ṁ_He × s_He

where, s_air and s_He are the specific entropies of air and helium respectively.s_mixture is the specific entropy of the mixture.Taking the reference temperature (T0) to be 295 K, the rate of entropy production can be calculated as:

S_gen = ṁ_mixture × [s_mixture − s_0(T0)] − ṁ_air × [s_air − s_0(T0)] − ṁ_He × [s_He − s_0(T0)]where, s_0(T0) is the entropy at reference temperature (T0).

s_mixture = 1.128 × ln [(0.21 × P_1)/(0.21 × P_1 + 0.79 × P_2)] + 0.881 × ln [(0.79 × P_2)/(0.21 × P_1 + 0.79 × P_2)] = 1.128 × ln [(0.21 × 1)/(0.21 × 1 + 0.79 × 1)] + 0.881 × ln [(0.79 × 1)/(0.21 × 1 + 0.79 × 1)] = 0.375 kJ/kg.K

where, P_1 and P_2 are the partial pressures of air and helium respectively at the exit temperature.Taking the specific entropies of air and helium from steam tables, we get;

s_air = 72 × 1.005 × [ln T3/T0] − 72 × R_air × [ln P1/P0] = 72 × 1.005 × [ln(573.8/295)] − 72 × 0.287 × [ln 1/1] = 349.1 J/kg.ks_He = 4.15 × 5.193 × [ln T3/T0] − 4.15 × R_He × [ln P2/P0] = 4.15 × 5.193 × [ln(573.8/295)] − 4.15 × 2.076 × [ln 1/1] = 602.1 J/kg.k

Where, R_air and R_He are the specific gas constants of air and helium respectively.The rate of entropy production (S_gen) = 76.15 × [0.375 − 0.365] − 72 × [0.349 − 0.300] − 4.15 × [0.602 − 0.176] = 0.767 kW/K

Therefore, the rate of entropy production is 0.767 kW/K.

(c) Calculation of the rate of energy destruction:

The rate of energy destruction is calculated using the equation:Energy destruction rate = S_gen × (T − T0)where, T is the temperature at which the entropy is generated.Substituting the given values, the rate of energy destruction is;Energy destruction rate = 0.767 kW/K × (573.8 K − 295 K) = 177.5 kW

Therefore, the rate of energy destruction is 177.5 kW.

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Answer: 75,000 Hz

Explanation:

To calculate the frequency of the supply voltage, we need to use the formula:

frequency = 1 / time period

where the time period is the time it takes for one complete cycle of the waveform. For a sine wave, the time period is given by:

time period = 2π / angular frequency

where the angular frequency is the rate of change of the angle with respect to time and is given by:

angular frequency = 2π × frequency

Comparing the given voltage expression v = 240 sin (471,239t) V to the general form of a sine wave:

v = Vm sin(ωt + θ)

where Vm is the amplitude, ω is the angular frequency, t is time, and θ is the phase angle, we can see that the amplitude is 240 V, the angular frequency is 471,239 rad/s, and there is no phase angle (θ = 0).

Using the formula for angular frequency, we can find the frequency:

angular frequency = 2π × frequency

471,239 = 2π × frequency

frequency = 471,239 / (2π) ≈ 75,000 Hz

Therefore, the frequency of the supply voltage is approximately 75,000 Hz.

Hope this helps, :^)

Answer:

 So its area A = 4 * 4sqrt(2) = 16sqrt(2) inches^2

Explanation:

In order to find the area of the section, we need to find the length of one of the diagonals.

Using the Pythagorean Theorem, a^2 + b^2 = c^2, we pick any side of the cube which in

itself is a square with sides 4 inches each.  The length of the diagonal of the square is

2(4^2) = c^2  or c = 4sqrt(2).

To calculate the area of the section, we must first determine the length of one of the diagonals. Using the Pythagorean Theorem, \(\bold{a^2 + b^2 = c^2}\), one selects any side of the cube which in its a square with four-inch sides.

           \(\to 2(4^2) = c^2 \\\\ \to c = 4\sqrt{(2)}\)

      \(\to A = 4 \times 4\sqrt{(2)} = 16\sqrt{(2)}\ inches^2\)

Therefore, the answer is "\(\bold{ 16\sqrt{(2)}\ inches^2}\)".

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The correct answer is  C) Foggy conditions made visibility poor.   statements about the fatal Tesla accident involving Joshua Brown is not true.

Fog reduces sight to one kilometre, making it impossible for you to see farther than that from where you are now standing. One to two kilometres of sight may be lost due to mist. Fog comes in a variety of forms, including freezing fog, valley fog, advection fog, and radiation fog. The small water droplets hanging in the air are to blame for the decreased visibility. In urban and industrial locations, where there are lots of pollution particles functioning as water droplet nuclei, the fog tends to be the thickest. Although fog may cause precipitation (drizzle), precipitation (drizzle) is not the same thing as fog. Fog, by definition, occurs when visibility is less than one kilometre, says the international definition (0.62 mile).

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Based on Harold's statement, it is most likely that he is trying to apply the element of 2. "Requirements Analysis" in systems engineering.

Customer Requirements Analysis means understanding what customers want and need from the system. It's important for systems engineering to make sure the system meets these needs and expectations. This means collecting and studying what the customers want so that the finished product will be what they need.

Harold says it's really important to figure out what the customers want, when they want it, and where they want it. This means he wants to focus on making customers happy and make sure the company is doing what customers want.

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Harold, is an systems engineer that works for a company involved with developing advanced condition based maintenance technologies for a variety of applications and industries. At a recent company strategy session Harold made the following statement in a meeting with his key managers: “We need to focus our attention on determining what our customers want, where they want it, and when they want it.” Based on this statement, Harold is most likely trying to apply what element of systems engineering?

1.

Needs Assessment.

2.

Requirements Analysis.

3.

Functional Allocation

4.

Design Synthesis

5.

Systems of Systems Integration

A sequence detector for detecting bit-pattern "010" can be implemented in Verilog using a finite state machine. The design requires three states: initial state, state 1, and state 2.

The input sequence is checked at each clock cycle and the state transitions occur based on the pattern detected. The Verilog code for the sequence detector can be written as a module with one input and one output. The input signal represents the binary sequence being monitored and the output signal is asserted high whenever the pattern "010" is detected. A finite state machine can be used to implement the functionality of the sequence detector.

The Verilog code for the sequence detector should be synthesized and simulated using a testbench on a simulator tool such as Edaplayground. com. The testbench should contain test cases that cover all possible combinations of the input signal. The simulation results should be verified to ensure the output signal is correct for each test case.

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The CIA triad and the Parkerian hex are the fundamental principles of information security.

Parkeriano, or Parkerian hexad: is a set of six elements of information security proposed by Donn B. Parker.

1. Confidentiality.

2. Ownership or Control.

3. Integrity.

4. Authenticity.

5. Availability.

6. Utility

The Parkerian hexagram adds three more attributes to the three classic security attributes of the CIA triangle

these are the fundamental principles of information security.

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This is true, because the IaaS providers refer to companies for engineering design that offer Infrastructure as a Service, which means they provide cloud-based infrastructure resources such as servers, storage, networking, and virtualization to their clients.

Responsible for the engineering design, setup, configuration, maintenance, and troubleshooting of all Windows Servers and related technologies. Amazon Web Services AWS and Rack space are both popular IaaS providers that offer these services to businesses and individuals. Therefore, the statement that IaaS providers are Amazon Web Services AWS and Rackspace is true.

Infrastructure as a Service IaaS providers are companies that offer cloud-based computing resources, such as storage, servers, and networking components, to customers on a pay as you go basis. Amazon Web Services AWS and Rack space are examples of IaaS providers, making the statement true.

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According to the given statement "tech a says that on short-/long-arm (sla) suspension systems, the upper control arm is the long one. tech b says that on sla suspension systems, the control arms are connected to the frame by ball joints," the correct answer is tech b.

Tech b is right about the fact that on sla suspension systems, the control arms are connected to the frame by ball joints.What is Short-/Long-Arm Suspension Systems (SLA)?Short-/Long-Arm Suspension Systems (SLA) is a front suspension system that comprises both upper and lower control arms. The upper control arm is longer than the lower control arm, and the two are connected to the frame by ball joints. When compared to the upper control arm, the lower control arm is shorter.The function of Short-/Long-Arm Suspension Systems (SLA) is to provide stability and improve vehicle handling, making it more comfortable to drive. The length of the upper control arm also aids in improving stability and handling. The lower control arm is shorter, allowing for the steering linkage and wheel assembly to be mounted further back for better weight distribution and balance.

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The given information in the question is as follows:

NA= 0.48k,

= 0.6k₂

= 1

Now, the formula for the minimum line width is given as follows:

Minimum line width = k₁λ/NA

where, k₁ = 0.6λ = wavelength

NA = numerical aperture

So, putting the given values in the above equation, we get:

Minimum line width

= (0.6 × λ)/0.48

= (5/4) × (k₁λ/NA)

= (5/4) × (0.6λ/0.48)

Minimum line width = 0.938 μm

Now, the formula for the depth of focus (DOF) is given as follows:

DOF = k₂λ/NA²

where, k₂ = 1λ = wavelength

NA = numerical aperture

So, putting the given values in the above equation, we get:

DOF = λ/NA²

DOF = λ/(0.48)²

DOF = 3.52 μm

Thus, the minimum line width is 0.938 μm and the depth of focus is 3.52 μm.

This wavelength is not suitable for current CMOS trends as the minimum line width required for current CMOS trends is much smaller than this value.

However, it is suitable for MEMS technology where the minimum feature size is generally larger than in CMOS technology.

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On average, you can find 2 carbon atoms in 100 unit cells in the Ferrite phase when the alloy is cooled to just below 727°C.

b) We need to consider that at temperatures below 727°C, plain carbon steel is in the Ferrite phase (α-Fe), which has a BCC (body-centered cubic) structure. In this phase, carbon solubility in iron is very limited, and the maximum solubility is approximately 0.022 wt%.

Since we are considering 100 unit cells, we will calculate the number of carbon atoms based on the reduced solubility:

1. Convert the 0.022 wt% solubility to atomic percent: (0.022 wt% C) / (0.022 wt% C + 99.978 wt% Fe) x 100 = 0.022 atomic % C
2. Calculate the number of carbon atoms in 100 unit cells: (0.022 atomic % C) x (100 unit cells) = 2.2 carbon atoms

Since we cannot have a fraction of an atom, we'll round this number down to the nearest whole number which is 2 carbon atoms.

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The evaluation method used for evaluating passive solar thermal systems is the Utilizability Method, and Heating Degree Days (HDD) can be determined using weather data and base temperature.

The Utilizability Method is a widely accepted approach for evaluating passive solar thermal systems' performance. It focuses on the ratio between the energy utilized and the total incident solar energy. This method takes into account the system's efficiency, as well as its ability to store and distribute heat.

To determine Heating Degree Days (HDD), you will need to gather weather data, specifically daily average temperatures. Choose a base temperature, which is typically 18°C (65°F) for buildings. For each day, subtract the daily average temperature from the base temperature. If the result is positive, it indicates a heating demand. Sum the positive differences over a specified period (e.g., a month or year) to calculate the total Heating Degree Days for that period.

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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.

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.

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Answer:

Explanation:

Please

Here we see that purchasing the computer is a better choice since the total cost of ownership over 15 years is less than the present value of leasing for the same period.

To determine the best option, we need to compare the present value of the cost of leasing with the present value of the cost of purchasing.

Option 1: Lease

Cost per day = $50

Number of days in a year = 365

Annual cost of leasing = $50/day × 365 = $18,250

Present value of annual leasing cost over 15 years at 9% interest rate:

PV(Lease) = $18,250 × [(1 - (1 + 0.09)^-15) / 0.09] = $173,186.76

Option 2: Purchase

Cost of computer = $25,000

Salvage value at the end of 15 years = $4,000

Annual maintenance cost = $2,800

Total cost of ownership over 15 years:

Total Cost = Cost of computer + Present value of annual maintenance cost over 15 years + (Cost - Salvage value) / Present value factor for 15 years

Total Cost = $25,000 + [$2,800 × ((1 - (1 + 0.09)^-15) / 0.09)] + [($25,000 - $4,000) / (1 + 0.09)^15]

Total Cost = $67,739.12

Comparing the two options, we see that purchasing the computer is a better choice since the total cost of ownership over 15 years is less than the present value of leasing for the same period. Therefore, management should choose to purchase the computer.

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The dark-side traits the leaders possessed that caused them to be ineffective are; Narcissism, Psychopathy and Machiavellianism.

According to Paulhus and williams, they discovered in year 2002 that several personality traits could be related to failures of a leader but they narrowed it down to three which they referred to as the Dark Triad.

These dark side traits are;

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Answer:

Explanation:

Technology is used in a variety of ways to improve casting products. Some examples include:

1.Computer-aided design (CAD) and computer-aided manufacturing (CAM) software: These tools allow for the creation of highly detailed and precise designs for casting products, resulting in improved product quality and consistency.

2.3D printing: This technology allows for the creation of complex, detailed molds for casting, as well as the ability to produce small quantities of parts without the need for traditional tooling.

3.Simulation software: Simulation software is used to predict the behavior of a casting product during the casting process, allowing for adjustments to be made to the design and manufacturing process to improve product quality and reduce defects.

4.Robotics and automation: Robotics and automation are increasingly used in casting operations to improve efficiency, reduce costs, and improve product consistency.

5.Metal Additive manufacturing: Metal additive manufacturing techniques such as Directed energy deposition, Electron beam melting, and Laser beam melting can be used to produce casting products with high precision and complex geometries, which are not possible with traditional casting methods.

6.Non-destructive testing techniques: Non-destructive testing techniques such as X-ray, ultrasonic, and magnetic particle inspection are used to detect defects and ensure product quality in casted products.

Overall, technology plays a crucial role in improving the casting process, allowing for more precise and efficient production of higher-quality products.

As a consequence of the properties of an ideal op-amp, the op-amp's input voltage will be **zero**.

An ideal op-amp has infinite voltage gain, which means it amplifies the voltage difference between its input terminals to an infinite value. Due to this characteristic, the input voltage of an ideal op-amp is considered negligible, as any voltage difference between the input terminals would be amplified to infinity.

In addition, an ideal op-amp has infinite input resistance, meaning it draws no current at its input terminals. Therefore, there is no voltage drop across the input terminals, resulting in an input voltage of zero.

It's important to note that while these properties hold true for an ideal op-amp, real-world op-amps have finite voltage gain, input resistance, and output resistance, which should be considered in practical circuit designs.

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Answer:

20% of the critical value. Determine (a) the damping factor, (b) the

Explanation:

Answer:

A)  222.58 kJ / kg

B)  0.8897 M^3/ kg

c)  0.7737 m^3/kg

D)  746.542 k

E)  536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) =  0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

 = Cv*T1 =  0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying  r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

\(\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }\)

hence T2 = \(9^{1.4-1} * 310\) = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718  * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

Answer:

a) Mt = 0.0023229

b) = U1 = 214.07

c) = V₁  = 0.861 m³/kg

d) = Vr1 = 621.2

Explanation:

Given that

R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4

specific heat at constant volume Cv = 0.7174 kJ/kg.K

Specific heat at constant pressure is 1.0045 Kj/kg.K

a)  To determine the total mass (kg) of air in the engine.

we say

P1V1 = mRT1

we the figures substitute

(100 x 10³) ( 500 x  10⁻⁶) = m ( 0.287 x  10³) ( 300 )

50 = m x 86100

m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)

Total mass of 4 cylinder

Mt = m x k

Mt = 0.0005807 x 4

Mt = 0.0023229

b) To determine the specific internal energy (kJ/kg) at state 1

i.e at T1 = 300

we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.

U1 = 214.07

c) To determine the specific volume (m³/kg) at state 1.

we say

V₁ = V1/m

V₁ = (500 x  10⁻⁶) / 0.0005807

V₁  = 0.861 m³/kg

d) To determine the relative specific volume at state 1.

To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.

At T1 = 300 k

Vr1 = 621.2

Answer:

Land capability is one aspect of land classification. For the determination of land capability, the usefulness of land for agriculture, forest and tourism is assessed solely on the basis of physical environmental factors.

Hence the capability denotes

The initial void ratio is the parameter which is used to show the structural foundations for each specimen of sand so that the method and speed of compression would be measured.

Relative density is the mass per unit volume of each specimen of sand which is measured and it has to do with the relative ratio of the density of the sand.

Unit weight is the the exact weight per cubic foot of the sand which is measured.

Please note that your question is incomplete so I gave you a general overview to help you better understand the concept

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