Find the second virial coefficient for the Dieterici equation of sta.pdf

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Find the second virial coefficient for the Dieterici equation of state p = RT/V-b exp (-a/RTV) Solution where a is associated with the interaction between molecules and b takes into account the finite size of the molecules, similarly to the Van der Waals equation. the viral coefficient are Tc= a/(4Rb) Pc= a/(4c2b2) Vc=2b.

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Did Carl Linnaeus incorporate evolutionary thinking in the development of his Systema Naturae? Solution Answer No Carl Linnaeus did not incorporated evolutionary thinking in the development of his Systema Naturae. He published this book in 1735 In this book Linnaeus presented an outline of the sexual system of classification of plant kingdom. He classified plant kingdom into 24 classes on the basis of presence and absence of stamens and their number, without taking into account any evolution relationship. Evolutionary thinking was incorporated into classifications after the publication of Darwin\'s Origin of Species (1859).

Did Carl Linnaeus incorporate evolutionary thinking in the developme.pdf

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Do you think wood is mechanically homogeneous (i.e. as strong in all directions? Why or why not? Solution Wood is mechanically heterogeneous because the composition of wood are of solid,some liquid and little gases when heated.There are differences in the grain that show the material is not uniformly mixed. This allows certain areas of the sample of wood to have different properties than other areas..

Do you think wood is mechanically homogeneous (i.e. as strong in all.pdf

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Design a database to store details about U.S. presidents and their terms in office. Also, record details of their date and place of birth, gender, and political party affiliation (e.g., Caluthumpian Progress Party). You are required to record the sequence of presidents so the predecessor and successor of any president can be identified. How will you model the case of Grover Cleveland who served nonconsecutive terms as president? Is it feasible that political party affiliation may change? If so, how will you handle it? In populating your solution include the following information (note this is not the format you will store it in the database): Chester Arthur Born 10/05/1829 Whig until 01/01/1854 Republican after that 21st president Grover Cleveland Born 03/18/1837 Democrat his entire career 22nd President 24th President Benjamin Harrison Born 08/20/1833 Whig until 01/01/1856 Republican after that 23rd President Chester Arthur Born 10/05/1829 Whig until 01/01/1854 Republican after that 21st president Grover Cleveland Born 03/18/1837 Democrat his entire career 22nd President 24th President Benjamin Harrison Born 08/20/1833 Whig until 01/01/1856 Republican after that 23rd President Solution Create table President( PresidentID integer(6) primary key, Name varchar (30) Place varchar (20), Dateofbirth date, Gender varchar(1) check in (m,f), Ruling_startdate date, Ruling_end date date, Affiliation_party varchar(20), Terms varchar(30)); We will design all the attributes under one table or define separate table for president personal details and party details. We can separate the tables now. President table: Create table president( PresidentId integer(6) primary key, Name varchar(30) foreign key, Dateofbirth date, Gender varchar(1) check in (m,f) Affiliation_party varchar(20), Terms varchar(20)); Create table pres_det( presidentID integer(6) primarykey, Place_of_elect varchar(20), Birthdate date, Ruling_startdate date, Ruling_enddate date, ); CREATE SEQUENCE seq_presidentID MINVALUE 10000 START WITH 10000 INCREMENT BY 100 Cycle. If we execute this sequence president id is set to default value and increment by 100 for every president value. DECLARE @maxTerm INT DECLARE @minTerm INT SELECT @maxTerm = MAX(term_id), @minTerm = MIN(term_id) FROM President AS p JOIN President AS p ON p. presidentId = pres_det.presidentId WHERE p.Name = \'name\' Then predecessor is SELECT p.Name FROM pres_det AS t JOIN President AS p ON p.presidentId = t.presidentId WHERE p.term = @minTerm - 1 To find the successor is SELECT p.Name FROM pres_det AS t JOIN President AS p ON p.presidentId = t.presidentId WHERE p.term = @maxTerm +1.

Design a database to store details about U.S. presidents and their t.pdf

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Chapter 1 - A View of Life and its Chemical Basis. 1. Explain the basic characteristics that are common to all living things. 2. Describe the levels of organization of life. 3. Distinguish among the three domains of life and kingdoms. 4. Explain the process of natural selection. 5. Identify the components of the scientific method. 6. Contrast ionic and covalent bonds. 7. Explain how the properties of water make life possible. 8. Recognize the importance of functional groups in determining the chemical properties of an organic molecule. 9. Summarize the categories of the main macromolecules and provide examples of their diverse biological functions: carbohydrates, lipids, proteins and nucleic acids. Solution please split the questions and ask. it was too lengthy to answer so all are in short. 1. Explain the basic characteristics that are common to all living things. The fundamental characteristics of all living beings are growth, reproduction. All are made up of cells, they adapt and respond to the environment. They all obtain and utilize energy. Though all these are in common, they have different cellular organization. 2. Describe the levels of organization of life. The various levels of organization of life include organelles, cells, tissues, organs, organ systems, organisms, populations, communities, ecosystem and biosphere. They are in the chronological order starting from the simplest form to the most complex one. 3. Distinguish among the three domains of life and kingdoms. Three domain system introduced by carl woese divides cellular forms into three domains- archae, bacteria and eukaryotes. It mainly separates archae from prokaryotes based on their 16sRNA differences. This classification is above the kingdom level. The five kingdom system has prokaryotes without membranes placed under Monera, eukaryotes are placed into 3 kingdoms plantae, fungi and animalia. Protista are a separate kingdom consisting of diversified algae and others which does not come under other groups 4. Explain the process of natural selection Charles Darwin proposed the natural selection in his theory of evolution. Natural selection is a process by which the organisms which are adapted to the environmental conditions and survive in them. Shortly said as “survival of the fittest”. They transfer the genetic traits to further generations making them more dominant while those who cannot survive get eliminated. 5. Identify the components of the scientific method. Scientific method is a planned method which helps scientists in answering or solving a problem. It consists of 6 components. 1. Purpose or question of research, 2. Research to find more information on the question, 3. Hypothesis after getting preliminary information, 4. Experimental design of the question, 5. Analysis of the results, 6. Conclusion of the experiment. 6. Contrast ionic and covalent bonds. Covalent bonds are less polar than ionic bonds.Covalent bond is a chemical bond formed between two non-metals .

Chapter 1 - A View of Life and its Chemical Basis. 1. Explain the.pdf

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CHOOSE THE CORRECT ANSWER: 1) PVC Corporation is considering an investment proposal in which a technology investment of $10,000 would be required. The investment would provide cash inflows of $2,000 per year for six years. The technology investment will have no salvage value at the end of six years. If the company\'s discount rate is 10%, the investment\'s net present value is closest to (Ignore income taxes.): (See the time value of money BELOW THE PAGE.) A. $1,289 B. $(1,289) C. $3,066 D. $2,000 2) RST Corporation has provided the following data concerning an investment project that it is considering: Initial investment $160,000 Annual cash flow $54,000 per year Salvage value at the end of the project $11,000 Expected life of the project years 4 years Discount rate 10% The net present value of the project is closest to: (See the time value of money BELOW THE PAGE) A. $175 B. ($23,694) C. $3,662 D. $11,175 3) An increase in the expected salvage value at the end of a capital budgeting project will A. Increase the cash outflows of the project. B. Increase the present value of cash inflows of the project C. Decrease the cash inflows of the project D. Decrease the net present value of a project HERE IS THE Time Value of Money Factor for the Present Value of a Sum PV =FV * TVM Factor Present Value of a Sum = FV * 1/ (1 + i) n 2% 3% 4% 5% 6% 8% 10% 12% 1 0.9804 0.9709 0.9615 0.9524 0.9434 0.9259 0.9091 0.8929 2 0.9612 0.9426 0.9246 0.9070 0.8900 0.8573 0.8264 0.7972 3 0.9423 0.9151 0.8890 0.8638 0.8396 0.7938 0.7513 0.7118 4 0.9238 0.8885 0.8548 0.8227 0.7921 0.7350 0.6830 0.6355 5 0.9057 0.8626 0.8219 0.7835 0.7473 0.6806 0.6209 0.5674 6 0.8880 0.8375 0.7903 0.7462 0.7050 0.6302 0.5645 0.5066 7 0.8706 0.8131 0.7599 0.7107 0.6651 0.5835 0.5132 0.4523 8 0.8535 0.7894 0.7307 0.6768 0.6274 0.5403 0.4665 0.4039 9 0.8368 0.7664 0.7026 0.6446 0.5919 0.5002 0.4241 0.3606 10 0.8203 0.7441 0.6756 0.6139 0.5584 0.4632 0.3855 0.3220 20 0.6730 0.5537 0.4564 0.3769 0.3118 0.2145 0.1486 0.1037 25 0.6095 0.4776 0.3751 0.2953 0.2330 0.1460 0.0923 0.0588 Time Value of Money Factor for the Present Value of an Annuity PV = Annuity * TVM Factor Present Value of an Annuity = X*((1-(1/(1+i)^n))/i) 2% 3% 4% 5% 6% 8% 10% 12% 1 0.9804 0.9709 0.9615 0.9524 0.9434 0.9259 0.9091 0.8929 2 1.9416 1.9135 1.8861 1.8594 1.8334 1.7833 1.7355 1.6901 3 2.8839 2.8286 2.7751 2.7232 2.6730 2.5771 2.4869 2.4018 4 3.8077 3.7171 3.6299 3.5460 3.4651 3.3121 3.1699 3.0373 5 4.7135 4.5797 4.4518 4.3295 4.2124 3.9927 3.7908 3.6048 6 5.6014 5.4172 5.2421 5.0757 4.9173 4.6229 4.3553 4.1114 7 6.4720 6.2303 6.0021 5.7864 5.5824 5.2064 4.8684 4.5638 8 7.3255 7.0197 6.7327 6.4632 6.2098 5.7466 5.3349 4.9676 9 8.1622 7.7861 7.4353 7.1078 6.8017 6.2469 5.7590 5.3282 10 8.9826 8.5302 8.1109 7.7217 7.3601 6.7101 6.1446 5.6502 20 16.3514 14.8775 13.5903 12.4622 11.4699 9.8181 8.5136 7.4694 25 19.5235 17.4131 15.6221 14.0939 12.7834 10.6748 9.0770 7.8431 HERE IS THE Time Value of Money Factor for the Pre.

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Civil Infrastructure Problems Why the public is not concerned about the condition of civil infrastructure while it affects them personally? Why the public do not comment more to their elected representatives? Solution Infrastructure is the backbone of a country and has serious impact on progress of that country and happiness of its people. The basic infrastructure comprises of transportation, communication, sewage, water and electric systems. Further elaborating, it covers roads, highways, bridges, telecommunication, ports, airports and railways. People use the infrastructure daily, in some form or other and are affected by it. Poor infrastructure does generate discontent and is reflected in people movements protesting and demanding rapid improvements. These protests are more visible in poor and developing countries where even the basic infrastructure of housing and allied services itself is deficient or non-existent. The protests are now visible in developed countries also where aging infrastructure because of lack of funds is deteriorating rapidly. In fact, it was a major point of agenda in recent presidential elections in USA. The administration is understood to be planning on a major overhaul of infrastructure costing trillion of dollars. In countries like India the Prime Minister personally monitors the overall progress of improvement of infrastructure as it was a major election promise on which he was elected. So, it is not correct to say that people do not complain of poor infrastructure,though it may be muted in some countries, cities , states and provinces .People do complain to their elected representative and that is why it is everywhere a major point of promises in the election agenda. These infrastructure projects tend to be high-cost investments; though they are vital to a country\'s economic development and prosperity. Therefore, their improvements take time and is generally spread over several years..

Civil Infrastructure Problems Why the public is not concerned about t.pdf

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Cell Structure and Function Label each of the arrows in the following slide image: AE Onion Root Tip: 1000X Solution A) Cells are in anaphase The cells are arranged such that the centromeres have split at this point and the chromatids have been separated by migrating to the opposite poles. B) Cells here are present in the early prophase of mitosis, where the centriole can now be seen to move towards the opposite poles of the cell. C) cell plate D) not clear.y distinguishable.

Cell Structure and FunctionLabel each of the arrows in the followi.pdf

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C Language ProblemThere are more than one wrong things I think.P.pdf

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Briefly discuss the adaptive changes seen in the chordates over the other animal phyla? Solution Animals in the phylum Chordata reveal four key features that appear at some point all through their development. a. Notochord - The chordates are called for the notochord: a flexible, rod-shaped design that is found within the embryonic stage of most chordates and also in the adult stage of some chordate species. It is found between the intestinal pipe and the nerve cord, providing skeletal help through along the body. In some chordates, the notochord acts while the primary axial support of the human body throughout the animal\'s lifetime. In vertebrates, the notochord is provide during embryonic development, of which time it induces the development of the neural pipe which acts as an assistance for the developing embryonic body. The notochord, however, is replaced by the vertebral column (spine) in most adult vertebrates. b.Dorsal Hollow Nerve Cord - The dorsal hollow nerve cord derives from ectoderm that sheets into a worthless pipe all through development. In chordates, it is found dorsally (at the top of the animal) to the notochord. In contrast to the chordates, other animal phyla are known by strong nerve cables which are positioned either ventrally or laterally. The nerve cord within most chordate embryos develops into the brain and spinal cord, which include the main anxious system. c.Pharyngeal Slits - Pharyngeal slits are openings in the pharynx (the place just posterior to the mouth) that expand to the surface environment. In organisms that are now living in aquatic surroundings, pharyngeal slits allow for the quit of water that enters the mouth all through feeding. Some invertebrate chordates use the pharyngeal slits to filtration food out of the water that enters the mouth. In vertebrate fishes, the pharyngeal slits develop into gill arches, the bony or cartilaginous gill supports. In most terrestrial animals, including mammals and birds, pharyngeal slits are providing just during embryonic development. In these animals, the pharyngeal slits develop into the jaw and inner ear bones. d.Post-anal Tail - The post-anal tail is a posterior elongation of the human body, increasing beyond the anus. The trail contains skeletal things and muscles, which give a way to obtain locomotion in aquatic species. In some terrestrial vertebrates, the trail also assists with stability, courting, and signaling when risk is near. In individuals and other apes, the post-anal tail is provide during embryonic development, but is vestigial as an adult..

Briefly discuss the adaptive changes seen in the chordates over the o.pdf

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Argue in favor or against the following statements (include examples, rationale, explanations, etc.): 1. Plants migrate 2. Plants defend themselves 3. Plants are good mothers. 4. Grasses are the most important food for humans. Solution Plant migration The Dispersal of the seeds is continuous process, by which plant species survives and spreads. Migration of plants is a result of the dispersal of the seeds directly to wider areas of earth,also a phenomenon constant in plant kingdom.Angiosperms species covered the whole earth migrating to the different countries along the different routes. Due to global warming,the plants are heading for hills for better survival. A study shows that 171 forest species in the Western Europe, among most of spp are shifting toward their a favourable locations to cooler spots for better survival.Plants are migrate toward higher altitudes only because of the climate change, most of world plant species diversity are found in the tropical areas, a very few studies, have examine either tropical mountain plant species are they affected by change in climate to same extent as the temperate plant species. new study has determined, major changes occurred in last two [2]centuries.If we compare the migration of the plant communities on Chimborazo volcano in the Ecuador with a historical data from the 1802, the Aarhus University researchers found that average upslope shifted more than [> 500 metres].and The entire boundary of vegetation have been moved upwards from the 4,600 metres to the 5,200 metres. The main region for this dramatically shift is only the climate change last 210 years 2.Plants defend themselves Plant can not run away from danger like most of the animal do,so that they have developed there own weapon & aroma & release toxic chemical in order to protect themselves, Plant are able to move anywhere, so that every single cell defend itself against the all attack from the invading pathogens, for example- fungi and bacteria. first line defense in plant The first line of defense in plants is intact & impenetrable barrier that composed of the bark & waxy cuticle. Both are protect plants against the herbivores. And Other adaptations against herbivores are include the, thorns the modified branches of plant, hard shells and the spines.[modified leaves]first line defense occur when the plant cells can detect the presence of the foreign invader like(flagella of the bacteria or chitin which found in fungi cell walls) and they alert the surrounding cells through releasing molecules that give massage to other cells so other cell ready to fight invader their defences. example, the alerted infected plant cells secrete molecules are harmful to invadors or plant cell build up their more cell walls for extra protection form invader. second line of defense is hypersensitive response (HR).infected plant cells around the infection site will kill [itself]themselves to stop pathogen to spreading infection throughout the plant. This pr.

Argue in favor or against the following statements (include examples.pdf

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27. What is a prosthetic group? Identify prosthetic groups in the following conjugated proteins: A prosthetic group is a part of the protein that is not composed of amino acids. Lipoproteins Carbohydrates Hemoproteins Lipids Flavin nucleotides Metalloproteins Glycoproteins Phosphate groups Fe, Ca, Copper, Zinc Phosphoproteins Flavoproteins Heme (iron porphyrin) Solution part-a: Hemoproteins contain hemoglobin, in which \"iron (Fe2+) is the prosthetic group useful to carry oxygen Part-b: Lipoproteins contain lipid moieties with double & triple bonds as prosthetic groups Part-c: Metalloproteins such as cobalamin contain \"cobalt\" as prosthetic groups in their structure of intrinsic factor human digestive system Part-c: Flavoproteins contain flavin as prosthetic group Glycoproteins is conjugate protein that contains \"glycosidic linkage\" as prosthetic groups.

27. What is a prosthetic group Identify prosthetic groups in the fol.pdf

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What is the role of sequence numbers in the rdt protocol? What is the role of timers in the rdt protocol? Solution The internet network layer provides only best effort service with no guarantee that packets arrive at their destination. Also, since each packet is routed individually it is possible that packets are received out of order. For connection-oriented service provided by TCP, it is necessary to have a reliable data transfer (RDT) protocol to ensure delivery of all packets and to enable the receiver to deliver the packets in order to its application layer. A simple alternating bit RDT protocol can be designed using some basic tools. This protocol is also known as a stop-and-wait protocol: after sending each packet the sender stops and waits for feedback from the receiver indicating that the packet has been received. Stop-and-wait RDT protocols have poor performance in a long-distance connection. At best, the sender can only transmit one packet per round-trip time. For a 1000 mile connection this amounts to approximately 1 packet (about 1500 bytes) every 20 ms. That results in a pathetic 75 KB per second rate. To improve transmission rates, a realistic RDT protocol must use pipelining. This allows the sender to have a large number of packets \"in the pipeline\". This phrase refers to packets that have been sent but whose receipt has not yet verified by the receiver..

What is the role of sequence numbers in the rdt protocolWhat is t.pdf

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Which of the following statements about mitosis is true? O a. Mitosis produces two daughter cells identical to the mother cell. b Mitosis is the division of the nuclear component (DNA) which is then followed by the division of the cytoplasm (cytokinesis). C.A mitotically dividing cell copies its DNA, separates these copies at opposite poles and then divides. d All of the above O e. A and B only Solution 1. Answer: a. Mitosis produces two daughter cells identical to the mother cell. Explanation- The chromosomes undergo duplication before mitosis. After this, identical chromosomes reach daughter cells. 2. Answer: b. During metaphase, chromosomes align at the metaphase plate, the equator of the mitotic spindle Explanation- The chromosomes align at the centre of the mitotic spindle and the two kinetochores of each chromosome get attached to the microtubules from opposite poles of the spindle. 3. Interphase- D. Cell growth. Replication of DNA. Prophase- G. The chromosomes condense, each with two sister chromatids. Mitotic spindle begins to form. Prometaphase- C. The nuclear envelope breaks down and the chromosomes attach to the mitotic spindle at their kinetochores. Metaphse- A. Chromosomes align at the equator of the mitotic spindle. For each chromosome, the sister chromatids are attached at the kinetochore to microtubules from opposing poles. Anaphase- E. The sister chromatids are pulled apart to opposite poles of the mitotic spindle. Telophase- F. A nuclear membrane forms around each set of chromosomes at the mitotic spindle\'s poles. Mitosis ends. Cytokinesis- B. Cell pinches in until cytoplasm is divided in two (in animal cells). Results in two daughter cells that each contain a nucleus. Explanation- Mitosis comprises of four basic phases, which according to their order are- prophase, metaphase, anaphase, and telophase. Prometaphase is the late phase of prophase. Cytokinesis is the division of the cytoplasm. Interphase is the preparatory stage of mitosis..

Which of the following statements about mitosis is true O a. Mitosis.pdf

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Which of the following factors would NOT contribute to allopatric speciation? The isolated population is exposed to different selection pressures than the ancestral population Different mutations begin to manifest in the gene pools of the separated populations The separated population is small, and genetic drift occurs. There is high gene flow between the two populations A population becomes geographically isolated from the parent population. Solution Ans: There is high gene flow between the tow population is not contribute to allopatric speciation. In allopatric speciation, a population splits into two geographically isolated populations due to formation of a barrier between portions of a population.The isolated populations then experience differentiating genotypic and phenotypic divergence as a result of different selective pressures in their differing environments. Additionally, genetic drift will occur differently, eventually differentiating the population’s genotypes and phenotypes. The separated populations will also experience different mutations that may persist differently in their respective environments. The each population have evolved differently to an extent that they remain isolated because changes are too great to enable genetic mixing through sexual reproduction. Allopatric speciation is not necessarily precluded if some individuals from one group cross the barrier and mate with members of the other group. In other words, allopatric speciation can occur if gene flow between groups is greatly reduced, but not entirely eliminated..

Which of the following factors would NOT contribute to allopatric spe.pdf

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What would be the exact Materials and Method for this experiment? 2. Material and methods 2.1. Subjects One hundred obese children (58 girls and 42 boys) were studied with a mean age (±SD) of 11.4 ± 2.7 (range 5–17 y). Obesity was defined as a body mass index (BMI) above the 97th percentile for age and sex according to the relevant national curves [28]. The participants were recruited as part of the pediatric medical consultations of the Pre- vention and Management Pediatric Obesity Network of Besançon (France). The study was conducted in accordance with the Declaration of Hel- sinki and approved by the local ethics committee (Comité de Protection des Personnes, Est-II, France, n°2015-A00763-46). All participants and their parents or legal guardians were fully informed of the experimental procedures and gave written informed consent. 2.2. Experimental procedures and measurements After a medical exam performed by a pediatrician to ensure that the participant had the ability to complete the whole protocol, the included subjects underwent different assessments on the same day: i) anthro- pometric characteristics and maturation status; ii) blood sample and blood pressure; iii) physical activity level; iv) sleep-disordered breathing. Solution 2. Material and methods: Assessment of anthro-pometric characteristics and maturation status in childhood who have obesity followed by examination of regular blood sample and monitoring blood pressure; monitoring of physical activity level as per regular ime scale, finally asking them for any presence of sleep-disordered breathing 2.1. Subjects - obese children --> One hundred obese children (58 girls and 42 boys) were studied with a mean age (±SD) of 11.4 ± 2.7 (range 5–17 y). Obesity was de-fined as a body mass index (BMI) above the 97th percentile for age and sex according to the relevant national curves.

What would be the exact Materials and Method for this experiment2.pdf

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What are the most challenges of wireless networks? Solution The most challenges in wireless networks are: 1. Security: Mobility of users increases the security threats in a wireless network. The present wireless networks utilize authentication and data encryption methods on the interface to provide security to the users. 2. Connectivity issues: The connectivity issues can be addressed in many ways such as, adding access points, dividing networks to sustain proper access, limiting the bandwidth of the guest network and implementing management and policy software. These issues may occur from the addition of applications to the network. 3. power and energy: A mobile device is generally useful, small in size, and committed to carry out a certain set of functions. When a device is permitted to move freely, it would usually be tough to receive a continuous supply of power. To preserve energy, a mobile device should be able to operate in a successful and well-organized manner..

What are the most challenges of wireless networksSolutionThe .pdf

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What is particle? What is a \"rigid body\"? What is \"conservative force\"? Give an example of a \"non-conservative\" force? The below relationship embodies what general principle of physical science? T_1 + V_1 = T_2 + V_2 What are the three ways in which a body can move? What is \"kinetics\"? What is \"kinematics\"? What is \"power\"? What is \"work\"? What significance does the date 07 December 1941 have in American history? Solution 2.rigid body is a body which does not under any deformation on application of load 3.conservative force is a force under which internal energy of system remains conserved 4.friction force is an example of non conservative force 6.a body can move either by pure translatory motion or pure rotary motion or with both rotary plus translatory 8.kinematics is the branch of mechanics which deals with motion of body without considering force 9.power is defined as rate of doing work 10.work is defined force multiplied fy displacement in the direction of applied load.

What is particle What is a rigid body What is conservative .pdf

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What are some of the obstacles or barriers to implementing EBP (evidenced based practices) in nursing? Provide a rationale for your answer. Since there are numerous topics on the issue, it is not appropriate to repeat one that has already been mentioned unless providing new information. Solution Ans: Obstacles or barriers to implementing EBP (evidenced based practices) in nursing includes (1) Evidence might have a limited role in decision-making processes; (2) aspects other than quality of care steer the evidence-based practice agenda; (3) some health care providers benefit less from evidence-based practice than others and (4) there is a lack of competences to put the evidence-based principles in practice. Policy makers might consider health care system characteristics from and strategies developed or suggested by others to respond to country-specific obstacles. Examples include but are not limited to; (a) providing incentives for patient-centred care coordination and patient communication, (b) supporting practitioners interested in applying research-related activities, (c) considering direct access systems and interprofessional learning to respond to the demand for autonomous decision-making from satellite professional groups, (d) systematically involving allied health professionals in important governmental advisory boards, (e) considering pharmaceutical companies perceived as \'the enemy\' an ally in filling in research gaps, (f) embedding the evaluation of evidence-based knowledge and skills in examinations (g) moving from (in)formative learning to transformative learning and (h) organizing high quality catch-up programs for those who missed out on evidence-based medicine in their curriculum..

What are some of the obstacles or barriers to implementing EBP (evid.pdf

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Type II hypersensitivity occurs when: a.An immediate allergic or anaphylactic type of reaction occur b. Antibodies arc formed against antigens on cell surfaces c. Antigen-antibody complexes arc deposited in tissues d. Reactions do not require antibody production Which of the following is an example of type III hypersensitivity: a. Anaphylaxis b. SLE- Systemic Lupus Erythematosus c. Transfusions reactions d. Transplant rejection. Which of the following is capable of establishing the MAC- membrane attack complex, effect cell lysis? a. Mast cells b. MHC I proteins c. MHC II proteins d. Complement proteins In Myasthenia Gravis a. Acetylcholine receptors on neurons arc bound by antibodies b. Acetylcholine receptors on the motor end plate arc bound by antibodies c. Acetylcholine vesicles arc bound antibodies d. An autoimmune response is not involved. Cytotoxic T cells (CD8^+) react to cells that haw: a. Foreign MHC class II proteins on their surfaces b. B cell receptors c. Foreign MHC class I proteins on their surfaces d. Soluble antigens Solution 13. b The type II hypersensitivity occurs when, the antibodies produced by the immune response bind to antigens on the patient\'s own cell surfaces. This causes a B cell response, because here antibodies are formked against the foreign antigen. 14. b SLE- Systemic lupus erythematosus, it is due to Nuclear antigens. It occurs when accumulation of immune complexes (antigen-antibody) takes place. Type III hypersensitivity causes inflammatory response and attraction of leukocytes. 15 d. complement proteins Transmembrane channels formed by membrane-attack complex (MAC) leads to cell lysis and death by disrupting the cell membrane of target cells. It involves proteins C5 through C9 16 b. Antibody block or destroy nicotinic acetylcholine receptors at the junction between the nerve and muscle. 17. c. Foreign MHC class I protein on thier surface. cytotoxic T cell or CD8+ T-cell or killer T cell are able to recognize a specific antigen. TCR to bind to the class I MHC molecule,.

Type II hypersensitivity occurs when a.An immediate allergic or ana.pdf

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The sum of two integers is 19. The larger integer is 5 less than twice the smaller. Find the two integers. Solution Let the larger integer be X and smaller integer be Y Given sum of two integers is 19 i.e., X+Y=19 The larger integer is 5 less than twice the smaller i.e., X=2Y-5 X+Y=19 By plugging X=2Y-5 in X+Y=19 we get 2Y-5+Y=19 3Y=24 Y=24/3=8 X+Y=19 X+8=19 X=19-8=11 Therefore the two integers are 11,8.

The sum of two integers is 19. The larger integer is 5 less than twi.pdf

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