Business Finance - 17742 Words

CHAPTER 9 THE COST OF CAPITAL (Difficulty: E = Easy, M = Medium, and T = Tough) Multiple Choice: Conceptual Easy: Capital components Answer: c Diff: E [i]. Which of the following is not considered a capital component for the purpose of calculating the weighted average cost of capital (WACC) as it applies to capital budgeting? a. Long-term debt. b. Common stock. c. Accounts payable and accruals. d. Preferred stock. Capital components Answer: d Diff: E [ii]. For a typical firm with a given capital structure, which of the following is correct? (Note: All rates are after taxes.) a. kd ke ks WACC. b. ks ke kd WACC. c. WACC ke ks kd. d. ke †¦show more content†¦Statements a and b are correct. e. Statements a and c are correct. Factors influencing WACC Answer: a Diff: E [ix]. Wyden Brothers has no retained earnings. The company uses the CAPM to calculate the cost of equity capital. The company’s capital structure consists of common stock, preferred stock, and debt. Which of the following events will reduce the company’s WACC? a. A reduction in the market risk premium. b. An increase in the flotation costs associated with issuing new common stock. c. An increase in the company’s beta. d. An increase in expected inflation. e. An increase in the flotation costs associated with issuing preferred stock. WACC and capital components Answer: c Diff: E [x]. Which of the following statements is most correct? a. The WACC is a measure of the before-tax cost of capital. b. Typically the after-tax cost of debt financing exceeds the after-tax cost of equity financing. c. The WACC measures the marginal after-tax cost of capital. d. Statements a and b are correct. e. Statements b and c are correct. WACC and capital components Answer: a Diff: E [xi]. A company has a capital structure that consists of 50 percent debt and 50 percent equity. Which of the following statements is most correct? a. The cost of equity financing is greater than or equal to the cost of debt financing. b. The WACC exceeds the cost of equity financing.Show MoreRelatedBusiness Finance1291 Words   |  6 PagesUBFF2013 BUSINESS FINANCE Question: 1. (a) Frodo Baggins has RM1,500 to invest. His investment counselor suggests an investment that pays no stated interest but will return RM2,000 at the end of 3 years. (i) (ii) What annual rate of return will Frodo earn with this investment? Frodo is considering another investment, of equal risk, that earns an annual return of 8%. Which investment should he make and why? (b) Samwise Gamgee was seriously injured in an industrial accident. He suedRead MoreFinance in Business542 Words   |  2 PagesThe term finance originated in France. The English society adapted the word in the eighteenth century to mean administration of money. It has become a permanent fixture of English dictionary since then. 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The Theories Of Child Development - 2240 Words

This assignment will discuss the main theories of child development and discuss the main principles across all area of development. The assignment will also evaluate an observation conducted on a child and link their behaviour to relevant child development theories. The assignment will firstly highlight some child development theories from behaviourists such as Bandura, humanists such as Maslow and constructivists such as Piaget. Secondly the child’s observation will be discussed and evaluated here relevant link to development theories will be made. Lastly the assignment will finish off with a reflection. Jennie Lindon defines development as† the sequence of physical and psychological changes experienced by human beings, which start with conception and continue throughout life.†(Jennie Lindon 2012:5) the early year’s foundation stage (EYTS) framework states that there are seven areas of learning and development however they view some are more important thes e are called the prime areas of development. There are three prime areas of development these are: communication and language, physical development, personal social and emotional development. The EYFS also states that â€Å"children develop in different ways at different times† (Statutory framework for the early years foundation stage 2014) which is why the development matters promote the message that every child is unique and has their own way of learning. 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Professional Nursing Practice Concepts and Perspectives †Solution

Question: Discuss about the Professional Nursing Practice for Concepts and Perspectives. Answer: Introduction: Nursing is a profession, where there professionals ate accountable for assessing, monitoring as well as evaluating the outcomes of the patient. Being a key member of multidisciplinary team, nurses need to identify the key health needs and treatment priority of the patient, on the basis of patient assessment (George, 2011). In this context, several theories and models has been developed, for guiding the nursing care process and identifying the key nursing care priorities. In this essay, the key focus is the 89 years old Australian immigrant patient Mrs. Barbara Green, who has recently diagnosed with some additional health and well being issues along with her past medical history. In this context, to identify the key nursing priorities through a systematic process, Levett-Jonesclinical reasoning cycle has been used (Levett-Jones, 2013). In addition to this theoretical framework, Millers functional consequences theory has also been used to identify impact of Mrs. Greens age upon her phy sical, mental, social and emotional health and wellbeing. The patient Mrs. Green is an 89 years old German immigrant, who is living in Australia since last 40 years. She is a retired school teacher. She was married to Henry, who died 2 years ago. Her son Dominik lives in Germany and visit her regularly, whereas Tracy, her daughter also lives throughout Australia, visit her regularly. Mrs. Green can speak both English and German. She started to participate in local Catholic Church and organize fundraising activities. She is respectful to their catholic culture and unwilling to leave her home and return back to Germany. Key age related changes observed in Mrs. Green is macular degeneration, for which she stopped driving, became unable to leave her house due to altered mobility. Isolation is another age related change; she lost physical contact with her friends and members of German Association. Other recently diagnosed age related changes diagnosed for Mrs. Green are constipation, weight loss, limited joint movement, occasional dizziness and painful joints. From her previous history, other age related deterioration of Mrs. Greens health status has been identified, which were rheumatoid arthritis and osteoarthritis. Risk factors identified for Mrs. Green include living alone in a single storey house, without any support. Another risk factor is gradual deterioration of her vision and mobility. Hyperthyroidism is another health risk associated with negative health consequences (Nieswiadomy, 2011). It has also revealed that her house is unusually unkempt and she has minimal amount of food. Therefore, risk for malnutrition, risk physical, chemical hazard as well as risk of fall has been identified. Isolation and reduced independence is increasing her risk of stress and depression. Negative functional consequences include reduced vision, mobility and isolation, i.e. she is unable go out from her home and join her German association. In addition, due to impaired mobility and vision, she is being unable to perform ADLs (Hunter, 2016). Mrs. Green has recently been identified with several age related negative functional consequences, which are identified as the age related changes, according to Millers functional consequences theory. For instance, osteoarthritis, reduced vision, reduced mobility, reduced ability to drive and perform her ADLs. It has also been identified that she has a wide social network, but due to these changes in her functionality, she is unable keep her commitments, making her isolated (Blais, 2015). She lives alone, in an unsupported condition. She refused to live her house and live with her son or daughter. However, her reduced functionality is making her worried for her further life. While processing the information gained from Mrs. Greens case study, it has been revealed that her reduced mobility and reduced vision are two key negative consequences of her aging process. These problems are contributing to the development of her third key concern, i.e. social isolation. Recently she has been diagnosed with severe symptoms of rheumatoid arthritis, macular degeneration as well as hypothyroidism. For instance, limited joint movement, pain and stiffness are the consequences of progressive rheumatoid arthritis and osteoarthritis, whereas reduced vision is the consequence of dry macular degeneration; on the other hand, non-significant weight loss is the consequence of hypothyroidism as well as malnutrition (Nieswiadomy, 2011). Analyzing the case of Mrs. Green, three key problems have been identified, based on which the nursing care priorities would be decided, goals will be established and care plan will be developed. The first key issue for Mrs. Green is her reduced mobility and gradually reducing vision. These two problems are reducing her ability to perform ADLs, ability to be connected with the German association or drive. In addition, she is also suffering from several severe consequences of arthritis and macular degeneration. The next problem identified for Mrs. Green is her non-significant weight loss and malnutrition. It has been identified that she is not eating properly, losing her weight significantly, along with occasional dizziness and memory impairment (Chitty Black, 2011). Thus, these problems are contributing significantly in her overall deterioration in health functioning. In addition, the third problem identified for Mrs. Green is her social isolation and loss of independence. From her c ase study, it has been revealed that she is highly connected with the social network, but due to her mobility and vision issues, she is no longer able to go out from her home and join her German association (Wold, 2013). It is making her isolated, socially excluded, which have a significant impact upon her mental health. In addition, she lives alone at her home. Thus, this problem should also be prioritized. After identifying three key problems for Mrs. Green, it is important to set goals for each problem, as it helps to develop appropriate care plan for the improvement of her heath and well being. The following table represents the goals set for each heath problem of Mrs. Green. Nursing care priority/ key problem Goal 1. Reduced mobility and gradually reducing vision 1. To improve physical movement 2. To improve visual function 3. To reduce risk of fall 4. To improve her ability to perform ADLs 2. Non-significant weight loss and malnutrition 1. To encourage healthy eating 2. To facilitate weight gain 3. Social isolation and loss of independence 1. To improve her social engagement 2. Improve her self-esteem and autonomy According the above mentioned goals, the next table will demonstrate appropriate care plan for Mrs. Greens health issue. Nursing care priority/ key problem Nursing intervention 1. Reduced mobility and gradually reducing vision 1. Physical exercise training- Mrs. Green can be referred to a physical therapist who can help her to guide through some joint movement exercise, for reducing joint stiffness, physiotherapy is also recommended for improving muscle flexibility and strength building. A physical exercise session including cardiovascular exercise would be helpful for muscle strengthening and improving her daily functioning (Reuben Tinetti, 2012) 2. Assistive technology- An advanced technology cane or walker can be provided with object detector. Her driving vehicle can be modified along with an advance driving training to improve her driving mobility. Her kitchen and other parts of house can be modified with assistive technologies to improve her daily performance (Gillespie et al., 2012). For improved vision, along with glasses, screen readers, screen enlargement application are recommended 3. hazard analysis- Hazard analysis should be done for reducing the risk of fall 2. Non-significant weight loss and malnutrition 1. Diet plan- A dietician can be appointed, who will make a diet chart for her, including protein intake, for weight gain and reducing of risk associated with hypothyroidism and malnutrition. 2. Health promotion and regular check up- A health promotion is required for enhancing her awareness towards her health status and importance of healthy diet. In this session, she will also be discussed about her risk of living alone at her home (Hall et al., 2011). She would be advice to appoint a care attendant, who can assist her in her daily functions. In addition, she would be motivated for visit her GP regularly and check her vital signs including thyroid hormone level. 3. Social isolation and loss of independence 1. Use assistive technology to meet German society- With the use of cane, wheelchair or modified vehicle, Mrs. Green can attend her German association, which will improve her mental status. In addition, with video conference tool, she can also be connected with her German society. It would eliminate the feelings of being isolated (Fairhall et al., 2012). 2. Appointing a care attendant- A care attendant can be appointed, who will assist her to perform her daily functioning and improve her physical and mental health, by reducing isolation, enhancing independence and reducing the risk of fall through elimination of any kinds of hazardous substances (Cameron et al., 2013). 3. Palliative care- As she has been isolated and being stressed, social support and palliative care is needed for helping Mrs. Green to cope with her age related changes and consequences. In this context, her cultural aspects would ne prioritized, to make her feel valued (Levett-Jones, 2013). Evaluating the entire procedure, i.e. identifying care priority, assessing and establishing goals for each priority and developing care plan, it has been revealed that each goals has critically linked with the care plans, aligning with three key nursing problems (Levett-Jones, 2013). Integrating these interventions in the nursing care plan will help achieve goals successfully. While dealing with Mrs. Greens care planning, I have learned how to identify actual nursing are priorities and develop appropriate care plan for individual patient, with the help of step by step process of clinical reasoning cycle. In addition, assessing the case study I have also revealed the importance of understanding and analyzing age related consequences, while dealing with an aged patient, with the help of Millers functional consequences theory. In this essay, the key focus was to review the process of client focused care plan development through the step by step process of clinical reasoning cycle. The case study in this essay has focused upon 89 years old women Mrs. Green, who is subjected to several age related health issues. These health consequences have been analyzed with the help of Millers functional consequences theory. Then, with the help of Levett Jones clinical reasoning cycle, the care priority of the patient has been identified and then according to these priorities goals has been established, which in turn helped in generating appropriate care plan for Mrs. Greens health and wellbeing. Reference List Blais, K. (2015).Professional nursing practice: Concepts and perspectives. Pearson. Cameron, I. D., Fairhall, N., Langron, C., Lockwood, K., Monaghan, N., Aggar, C., ... Kurrle, S. E. (2013). A multifactorial interdisciplinary intervention reduces frailty in older people: randomized trial.BMC medicine,11(1), 65. Chitty, K. K., Black, B. P. (2011). Professional nursing: concepts challenges. Fairhall, N., Sherrington, C., Kurrle, S. E., Lord, S. R., Lockwood, K., Cameron, I. D. (2012). Effect of a multifactorial interdisciplinary intervention on mobility-related disability in frail older people: randomised controlled trial.BMC medicine,10(1), 120. George, J. B. (2011).Nursing Theories: The Base for Professional Nursing Practice, 6/e. Pearson Education India. Gillespie, L. D., Robertson, M. C., Gillespie, W. J., Sherrington, C., Gates, S., Clemson, L. M., Lamb, S. E. (2012). Interventions for preventing falls in older people living in the community.Cochrane Database Syst Rev,9(11). Hall, S., Kolliakou, A., Petkova, H., Froggatt, K., Higginson, I. J. (2011). Interventions for improving palliative care for older people living in nursing care homes.The Cochrane Library. Hunter, S. (Ed). (2016). Millers nursing for wellness in older adults (2 nd Australia and New Zealand ed.) North Ryde, NSW: Lippincott, Williams and Wilkins. Levett-Jones, T. (Ed.). (2013). Clinical reasoning: Learning to think like a nurse. Frenchs Forest, NSW: Pearson. Nieswiadomy, R. M. (2011).Foundations in nursing research. Pearson Higher Ed. Reuben, D. B., Tinetti, M. E. (2012). Goal-oriented patient carean alternative health outcomes paradigm.New England Journal of Medicine,366(9), 777-779. Wold, G. H. (2013).Basic geriatric nursing. Elsevier Health Sciences.

Specific heat of solids Essay Example For Students

Specific heat of solids Essay The objective of the study is to explain, measure and better understand the specific heat of copper and lead using the method of mixtures. Heat is a form of energy it is either expressed in joules, calories, or kilo-calories According to the law formulated by the French chemists Pierre Louis Dulong and Alexis Thrse Petit, the specific heat of solids which is characterized as the amount of heat required to raise the temperature of one gram of a substance to one degree Celsius specimens are inversely proportional to their atomic weights; that is, the specific heat multiplied by the atomic weight is approximately a constant quantity for all the solid elements. (http://encarta.msn.com). The heat capacity C of an object is defined as c= Q/m?T, where Q is the amount of heat required to change the temperature of the object by T. The specific heat c of a substance is the heat capacity per unit mass. The specific heat is measured in J/kgoC or cal/goC or kcal/kgoC. Suppose we have two objects, one hot and one cold. Let m1 and m2 be the masses of the hot and cold objects,T1 and T2 be the temperatures of the hot and cold objects, a nd c1 and c2 be their specific heats respectively. These two objects are brought into thermal contact with each other and allowed to reach a common final equilibrium temperature T3. We are assuming the system to be thermally insulated from the surroundings. According to conservation of energy, the heat gained by the cold object would equal the heat lost by the hot object. For this experiment, consider your system to consist of mixing a given mass m1 of a ?hot specimen with specific heat c1 at temperature T1 and a known mass m2 of water with specific heat c2 at a lower temperature T2 contained in a calorimeter of mass m3 with specific heat c3 also initially at temperature T2. Once again, we assume the system to be thermally insulated from the surroundings, and the heat capacity of the thermometer, which records the temperature, can be neglected. Let the final temperature of the mixture be T3. Energy conservation gives:Qlost(specimen) = Qgained(Water) + Qgained(Calorimeter)which yield s the unknown specific heat c1 of the specimen asWe assume that the mixing can be done without loss of heat by the hot specimen to the surroundings. We will write a custom essay on Specific heat of solids specifically for you for only $16.38 $13.9/page Order now We will consider a specimen heated to a high temperature is dropped into water contained in a calorimeter cup at a lower temperature. If this system is thermally insulated from the surroundings, the specific heat of the specimen can be determined by equating the heat lost by the metal to the heat gained by both the calorimeter cup and the water contained in it. (http://www.physics1.howard.edu/MSIP/GenLab1/GL1-10.pdf )The initial masses of boiler dipper, copper, lead, dipper and shots, inner vessel with water, and water are measured and recorded. The specimens (copper/lead) are heated and the initial temperatures are recorded after which, it is dropped in the calorimeter containing cold water. The temperature rise of the water in the calorimeter is observed and recorded. From the data gathered, the specific heat of the specimen can be known through the formula stated above. Mass of Dipper and Shots,m2 (gm)320280Mass of inner vessel w/ water,m3 (gm)220220Mass of water, mw=m3-mc (gm)165165Temperature of hot shots,T1 (oC)8384Initial Temperature of System, T2 (oC) 2221Final temperature of system,T3 (oC)2626Specific heat of water,cw (cal/gmoC)1.001.00Specific heat of specimen,cw (cal/gmoC)(Experimental)0.0520.072Specific heat of specimen (cal/gmoc) (standard)0.0360.092 Qlost(specimen) = Qgained(Water) + Qgained(Calorimeter) ms(T1 ? T3) cs = ((1)(165)+(0.215)(55))(26-21)% ERROR = / Standard ? Experimental/ x 100% StandardBased on the data above, I can say that the specific heat of material differ depending on the kind of substance. The specific heat for copper for example, after the experiment, became 0.052 cal/gmoC while on the other hand; the lead?s specific heat became 0.072 cal/gmoC. The 2 solids have achieved different results even though the procedure done for both was the same. This is because the different specimens differ in tolerance when heat is c oncerned. Therefore, the required amount of heat needed to change the temperature of a unit mass of substance by one degree also varies. .u51ef9fb970a7cfcc1bd98dad5093ba2a , .u51ef9fb970a7cfcc1bd98dad5093ba2a .postImageUrl , .u51ef9fb970a7cfcc1bd98dad5093ba2a .centered-text-area { min-height: 80px; position: relative; } .u51ef9fb970a7cfcc1bd98dad5093ba2a , .u51ef9fb970a7cfcc1bd98dad5093ba2a:hover , .u51ef9fb970a7cfcc1bd98dad5093ba2a:visited , .u51ef9fb970a7cfcc1bd98dad5093ba2a:active { border:0!important; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .clearfix:after { content: ""; display: table; clear: both; } .u51ef9fb970a7cfcc1bd98dad5093ba2a { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .u51ef9fb970a7cfcc1bd98dad5093ba2a:active , .u51ef9fb970a7cfcc1bd98dad5093ba2a:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .centered-text-area { width: 100%; position: relative ; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .u51ef9fb970a7cfcc1bd98dad5093ba2a:hover .ctaButton { background-color: #34495E!important; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .u51ef9fb970a7cfcc1bd98dad5093ba2a .u51ef9fb970a7cfcc1bd98dad5093ba2a-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .u51ef9fb970a7cfcc1bd98dad5093ba2a:after { content: ""; display: block; clear: both; } READ: Why do people help others? EssayAlthough the method mixture is an effective way to measure the specific heat of a given specimen, it is not entirely accurate. The errors in the experiment were primarily due to outside factors such as the wind, weather, etc. Also, the stirrer of the calorimeter added to the mass of the calorimeter which provided inaccuracy in measurement. In conclusion, I can say that the method of mixtures is a simple but an effective way to compute for the specific heat of specimens. In addition, the experiment helped clarify the concept of specific heat. I have learned that the specific heat is the amount of heat required by an object for it to change its temperature by a certain amount usually by one degree. This was shown when the specimens were transferred from one container to another that differed in temperature.1. The thermometer also gains some of the heat. Neglecting the heat capacity of the thermometer, it causes inaccuracy in the measurement of specific heat. Cs = ((200)(1)+(100)(0.092))(22-20))Bibliography: