Answer: I personally believe it does, because the Circle of Life, or what I think, at least buying and selling
Explanation:
Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Answer:
p q output ¬(p ∧ q)
0 0 1
0 1 1
1 0 1
0 0 0
p q output ¬p ∨ ¬q
0 0 1
0 1 1
1 0 1
0 0 0
Explanation:
We'll create two separate truth tables for both sides of the equation, and see if they match.
The expressions in the question use AND, OR and NOT operators.
The AND operation needs both inputs to be 1 to return a 1.The OR operation needs at least 1 of the inputs to be 1 to return a 1. The NOT operation takes a 1 and turns it into a 0, or takes a 0 and turns it into a 1.Let's start with ¬(p ∧ q)
NOT (0 AND 0) = NOT (0) = 1NOT (0 AND 1) = NOT (0) = 1NOT (1 AND 0) = NOT (0) = 1NOT (1 AND 1) = NOT (1) = 0Now let's move on to the second expression ¬p ∨ ¬q
NOT(0) OR NOT(0) = 1 OR 1 = 1NOT(0) OR NOT(1) = 1 OR 0 = 1NOT(1) OR NOT(0) = 0 OR 1 = 1NOT(0) OR NOT(0) = 0 OR 0 = 0Therefore we can say the two expressions are equivalent.
Attached the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:
What is the explanation of the truth table?As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.
De Morgan's law is a fundamental principle in propositional logic.
It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.
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While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
1. Is the team developing confidence that the solution space has been fully explored?
2. Are there alternative diagrams and alternative ways to decompose the problem?
3. Have external sources been thoroughly pursued, and everyone’s ideas been accepted and integrated in the process?
4. All of the above
Answer:
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
4. All of the above
Explanation:
The team must explore its solution space, including some external sources. Then, it must integrate its findings with the ideas of team members, ensuring the consideration of all possible ways to decompose the problem. This is because employing a structured process to concept generation enables the team to come up with creative solutions to design concepts.
What are three types of land reform
Answer:
Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);
Types of Land Reform
Abolition of Intermediaries
The first step taken by the Indian government under land reforms post-independence was passing the Zamindari Abolition Act. The primary reason of a backward agrarian economy was the presence of intermediate entities like, jagirdars and zamindar who primarily focussed on collecting sky-rocketing rents catering to their personal benefits, without paying attention to the disposition of farms and farmers. Abolition of such intermediaries not only improved conditions of farmers by establishing their direct connection with the government but also improved agricultural production.
Regulation of Rents
This was in direct response to the unimaginably high rents which were charged by intermediaries during British rule, which resulted in a never-ending cycle of poverty and misery for tenants. Indian government implemented these regulations to protect farmers and labourers from exploitation by placing a maximum limit on the rent that could be charged for land.
Tenure Security
Legislations were passed in all states of the country to grant tenants with permanent ownership of lands and protection from unlawful evictions on expiry of the lease. This law protects tenants from having to vacate a property immediately after their tenure is over unless ordered by law. Even in that case, ownership can be regained by tenants with the excuse of personal cultivation.
Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend auaching fins both inside and outside the tubes?
Answer:
Fins should be attached outside the tube Fins can be attached on both sides when convection coefficient of air inside the tube is equal to the convection coefficient of atmospheric air outside the tubeExplanation:
The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.
The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ), BUT
When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )
lists at least 6 units of measuring atmospheric pressure
Answer:
On my console displays for the ISS visiting vehicles, three units are used. The Americans use pounds per square inch (psi). The Russians use kilopascals (kPa). The Japanese use Torr - millimeters of mercury (mmHg). A fourth unit is simply the atmosphere, or multiples of it. So, for example, sea level air pressure (which is what we use onboard ISS) is defined as 1 atmosphere. That is equivalent to 14.7 psi, 101.3 kPa, or 760 mmHg.Here N represents newton which is SI unit of Force which is same as Kg.m/s2." role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2.2.
m represent metre which is SI unit of length.
Kg represent Kilogram which is SI unit of Mass.
m2SI" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 15px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">2SI2SI unit of Area.
Hope it helps.
Thanks.
Answer:
Pounds per square inch (psi)
Kilopascals (kPa)
Millimeters of mercury (mmHg)
Pascal (Pa)
Megapascal (MPa)
Atmospheric pressure (atm)
Hope this helps!
Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh xe AB = L ( m) . Ô tô không bị trượt ngang và đang quay vòng trên đoạn đường có góc nghiêng aphal , bán kính quay vòng r ( m ), vận tốc xe v ( m/s ). Tính chiều cao trọng tâm lớn nhất để xe không bị lật ngang .
Answer:
wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh
Explanation:
jwhwhwhwhwhwwhhahwhahahwh
Theo Anh / Chị, để đáp ứng yêu cầu phát triển nền kinh tế thị trường định hướng Xã hội Chủ nghĩa ở Việt Nam trong bối cảnh thời đại hiện nay, cần chú trọng giải quyết những vấn đề gì ?
How many numbers multiple of 3 are in the range [2,2000]?
Trình bày sự khác nhau của Dây chuyền đẳng nhịp đồng nhất, dây chuyền đẳng nhịp không đồng nhất, cho ví dụ minh họa
Functional and nonfunctional requirements documents are used to _____.
define the financial budget of a system
define the purpose of a system
facilitate communication between the users and a system
help exercise control over the inner workings of the firm.
Answer:
Non-functional requirements when defined and executed well will help to make the system easy to use and enhance the performance
Explanation:
Uses of P-N junction
Answer:
Explanation:
Two that come to mind:
a semiconductor diode is essentially a PN junctiona transistor is made of two pn junctions.Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe
how he would plan and manage projects in the future?
Select an answer:
project highlights
major changes and risks
summary of schedule and cost performance
summary of project management effectiveness
Answer:
Dalton
The part of the close-out report that would describe how he would plan and manage projects in the future is:
summary of project management effectiveness
Explanation:
The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.
State three types of maintenance.
Answer:
Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:
Explanation:
Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.
Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.
Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.
Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.
Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Sieve analysis Sieve Size No. 4 (4.75 mm) No. 10 (2.00 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Percent passing by weight 80 60 30 10 Atterberg limits Liquid limit (LL) Plastic limit (PL 31 25
(a) Classify this soil according to USCS system, providing the group symbol for it. Show how you arrive at the final classification.
(b) According to USCS system, what is a group name for this soil?
(c) Is this a clean sand? If not, explain why.
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
A) Classifying the soil according to USCS system
( using 2nd image attached below )
description of sand :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Several applications are listed below. Determine the relative importance of the resilience and toughness of the materials chosen for each application. Sort the items based on whether resilience is most important, toughness is most important, or both are equally important.
a. a non-critical spring that is used repeatedly
b. a high tension music wire
c. a one use safety device that absorbs impact energy
d. a burst panel designed to rupture at
e. a certain pressure
d. a structural member in a building
Answer:
they are important together, but if you want to use just one future you must think about which one is first needed. and then try to learn for economical so don't use more money
g A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120- Vrms, 60-HZ AC. (a) When both motors working together what is combined power factor
complete Question
A motor driving a 1000-W water pump has a power factor of 0.80 lagging; a second motor driving a 600-W water pump has a power factor of 0.60 lagging assuming the motors are working under 120-Vrms, 60-HZAC. When both motors working together what is the combined power factor? If a 200-μF capacitor is connected to the above system (two motors) what is the new combined power factor?
Answer:
[tex]p.f'=0.960[/tex]
Explanation:
First motor Power [tex]P=1000W[/tex]
First motor Power factor [tex]P.f=0.80[/tex]
Second motor Power [tex]P=600W[/tex]
Second motor Power factor p.f=0.60
Voltage [tex]V=120Vrms[/tex]
Frequency [tex]F=60Hz[/tex]
Capacitor [tex]C=200\mu F[/tex]
Generally power in Var is given as
For First Motor
[tex]Q=\frac{1000}{0.8}\sqrt{1-0.8^2}[/tex]
[tex]Q=750Var[/tex]
For Second Motor
[tex]Q=\frac{600}{0.6}\sqrt{1-0.6^2}[/tex]
[tex]Q=800Var[/tex]
Generally the equation for The Reactive Power is mathematically given by
[tex]Q_c=\frac{V^2}{X_c}[/tex]
Where
[tex]X_c=\frac{1}{2 \pi fc}[/tex]
[tex]X_c=\frac{1}{2 \pi 60*200*10^{-6} }[/tex]
[tex]X_c=13.3[/tex]
Therefore
[tex]Q_c=-\frac{120^2}{13.3}\\\\Q_c=-1085.97j[/tex]
Giving
Total Power Drawn by Supply
[tex]P_t=(1000+j750)+(600+800)-j1085.97[/tex]
[tex]P_t=1600+464.03j[/tex]
Therefore
[tex]p.f'=\frac{1600}{\sqrt{1600^2+464.03^2}}[/tex]
[tex]p.f'=0.960[/tex]
Suppose that you have a block of copper which is 300 g. Relevant material parameters follow.
Material atomic weight Density
Copper 63.5 g/mol 8.96 g/cm^3
Using equipartition, compute the molar heat capacity in J/mol K.
Answer:
[tex]C_v= 24.942[/tex] J / mol K
Explanation:
Molar Heat Capacity
using the equipartition, the equation :
[tex]$C_v=\frac{d}{2} R$[/tex]
Here, [tex]C_v[/tex] = molar heat capacity
d = degree of freedom
R = Universal gas constant
Degree of freedom, d is 6
Universal gas constant, R = 8.317 J/ mol K
Therefore, the molar heat capacity is :
[tex]$C_v=\frac{d}{2} R$[/tex]
[tex]$C_v=\frac{6}{2} \times 8.314$[/tex]
[tex]C_v= 24.942[/tex] J / mol K
three forces equal to 3p,5p,7p act simultaneously along the three side AB, BC,and CA of equilateral triangle ABC of side a. find the magnitude, direction and position of the resultant.
The given parameters include;
side AB = 3pside BC = 5p side CA = 7pFrom the image uploaded;
The resolution of the vectors along the sides of the triangle is as follows;
the y-component = force x sin(θ)the x-component = force x cos (θ)forces -------angle (θ) --------y-component-------x-component
3p ------------- 60⁰ --------- ----2.598 p ------------------- 1.5 p
5p -------------- 60⁰ ------------ (-4.33p) ------------------- 2.5 p
7p -------------- 60⁰ ------------- 0 -------------------------- (-7p)
sum: ( -1.732 p) y (-3P)x
The resultant of the vectors:
R² = (-1.732 p)² + (-3P)²
R² = 11.998 p²
R = √(11.998 p²)
R = 3.46 p
The direction of the vector:
[tex]\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{-1.732}{-3} )\\\\\theta = tan^{-1} (0.577)\\\\\theta = 29.98 \ ^0[/tex]
The position of the vector = 180 - θ = 180 - 29.98 = 150⁰ (second quadrant)
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định khoản nghiệp vụ sau : tạm ứng cho nhân viên A đi công tác bằng tiền mặt 50.000
Answer:
report on a fight you have witnessed
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/tex]
A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the water velocity is 3 m/s, how high will the water rise in the vertical leg relative to the level of the water surface of the stream?
Answer:
[tex]h=0.46m[/tex]
Explanation:
From the question we are told that:
Velocity of water [tex]V=3m/s[/tex]
Height=?
Generally, the equation for Water Velocity is mathematically given by
[tex]V=\sqrt{2gh}[/tex]
Therefore Height h is given as
[tex]h=\frac{v}{2g}[/tex]
[tex]h=\frac{3^2}{2*9.81}[/tex]
[tex]h=0.46m[/tex]
1) (30 pts ) Oxygen (O2) flows through a pipe, entering at at 4 m/sec at 10000 kPa, 227oC. For a pipe inside diameter of 3.0 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas
Complete Question
Nitrogen (N2) flows through a pipe, entering at at 4 m/sec at 1000 kPa, 2270C. For a pipe inside diameter of 3 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas Then using your ideal gas mass flow rate find the rate at which enthalpy enters the pipe (kJ/sec) NO Cp, Cv, k permitted
Answer:
[tex]H=9.91kJ/sec[/tex]
Explanation:
From the question we are told that:
Velocity [tex]v=4 m/sec[/tex]
Pressure [tex]P=1000kPa[/tex]
Temperature [tex]T=227 \textdegree C[/tex]
Diameter [tex]d=3cm=>0.03m[/tex]
Generally the equation for volumetric Flow Rate is mathematically given by
[tex]V_r=(\frac{\pi*d^2}{4}v)[/tex]
[tex]V_r=(\frac{\pi*(0.03)^2}{4} *4)[/tex]
[tex]V_r=0.002827m^3/s[/tex]
Generally the equation for mass Flow Rate is mathematically given by
[tex]m_r=\frac{PV_r}{RT}[/tex]
[tex]m_r=\frac{1000*0.002827}{0.297*(227+273)}[/tex]
[tex]m_r=0.019kg/sec[/tex]
Generally the equation for mass Flow Rate is mathematically given by
Using gas Table for enthalpy Value
[tex]T=500K=>h=520.75kg[/tex]
Therefore
[tex]H=mh[/tex]
[tex]H=0.019*520.75[/tex]
[tex]H=9.91kJ/sec[/tex]
which type of clectrical circuit is represented by this diagram?
Answer:
parallel
Explanation:
All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance it slides in 10seconds if the kinetic coefficient is 0.4.
Answer:
21 m
Explanation:
Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.
Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N
The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N
The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N
If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.
Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.
For the block to be in contact with the surface, the vertical forces on the block must balance.
Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,
N = mg + F₃" (since both the weight and the resultant vertical force act downwards)
N = mg + F₃"
Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N
So,
N = mg + F₃"
N = 35 kg × 9.8 m/s² + 8.82 N
N = 343 N + 8.82 N
N = 351.82 N
So, the net horizontal force F = F₃' - f.
F = 155.41 N - 0.4 × 351.82 N
F = 155.41 N - 140.728 N
F = 14.682 N
Since F = ma, where a = acceleration of block,
a = F/m = 14.682 N/35 kg = 0.42 m/s²
To find the distance the block moved, x we use the equation
x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²
Substituting the values of the variables into the equation, we have
x = ut + 1/2at²
x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²
x = 0 m + 1/2 × 0.42 m/s² × 100 s²
x = 0.21 m/s² × 100 s²
x = 21 m
So, the distance moved by the block is 21 m.
1025 steel wire is stretched with a stress of 70 MPa at room temperature 20 C. If th length is held constant, to what temperature in 'C and 'F must the wire be heated to reduce the stres to 17 MPa?
Check attached pictureCheck attached pictureCheck attached pictureCheck attached picture
Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address
lessons learned on a project?
Select an answer:
Add lessons learned as a topic in status meetings
Review past lessons learned so a new one does not have to be created,
Create lessons learned at the end of the project.
Brainstorm lessons learned at the beginning of a project
Answer: Create lessons learned at the end of the project.
Explanation:
Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.
The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.
On a ship the price gift is 24 euros .What is the difference in the price on a day when the exchange rate is £1=2378
Answer:
The answer is "[tex]\$29.7072[/tex]".
Explanation:
Please find the complete question in the attachment file.
As the original prices are 24 euros, [tex]\$ 1.2378[/tex] must be multiplied by 24.
[tex]\to 1.2378 \times 24 = \$29.7072[/tex]
Therefore the new price value will be [tex]\$ 29.7072[/tex]
beacuse thye want them to hav egoood and thye wn thme tto
Answer:
I don't understand the question
. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.
Answer:
not understand language
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold reservoir at TC = 400 K. For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
a.QH = 900 kJ, Wcycle= 450 kJ
b. QH = 900 kJ, Qc = 300 kJ
c. Weycle = 600 kJ, Qc= 400 kJ
d. η = 75%
Answer:
a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.
Explanation:
Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:
[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)
Where:
[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.
The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])
[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{r} = 66.667\,\%[/tex]
In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:
[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)
Where:
[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.
[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.
A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].
Now we proceed to solve for each case:
a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]
[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 50\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]
[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 66.667\,\%[/tex]
Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.
c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]
[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 60\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.
