A Research Methodology On Marketing Strategy And Motivation

4.0. Methodology This section explores the research methodology used. It is split into 5 segments. These include the research approach, design, data collection, analysis and ethical considerations. To summarise, the research takes an interpretive approach to understanding internationalisation strategy and motivation. This section outlines the use of a qualitative research design, implementing non-probability sampling to gather information from the sampling units. Data was collected using in-depth interviews with international marketing managers (sampling unit) that represented each of the institutions. These interviews have been transcribed – See Appendix 1. Once transcribed, data was analysed using open coding, in accordance with Strauss and Corbin’s (1997) grounded theory. Finally, the methodology considers ethical issues related to anonymity of each institution. 4.1. Research Approach Saunders et al (2009) define paradigm as a process of understanding and identifying social phenomena. Burrell and Morgan (1979) developed a framework to understand the four major paradigms of social theory. These include, radical humanist, radical structuralist, interpretive and functionalist – See Figure 3. Figure 3: The four paradigms of social theory (Saunders et al, 2009, p.120) The bottom-left segment represents the interpretive paradigm, influenced by the subjectivist and regulative conceptual dimensions. Subjectivism is concerned with understanding the meanings associatedShow MoreRelatedThe Radio And Media Company1133 Words   |  5 Pagesexperimenting different way of doing the marketing strategies, because it is one of the best way we can learn to get better and also i got to do it on a live dynamic market it has given me a lot of input about the do s and don ts on the marketing business. 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Roman and Early Christian Architecture Free Essays

Explain the techniques used by roman architects and describe the types of buildings they created. Use examples to support your essay. The building techniques the Roman architects used were the arches and domes. We will write a custom essay sample on Roman and Early Christian Architecture or any similar topic only for you Order Now They were very skilled at building with bricks and stones. These materials helped them develop new roof forms. To create arches they used wedge-shaped stones or bricks called vigorous. They placed the narrow ends on the inside of the arch and the wide ends on the outside. The Romans typically used round or pointed arches to span openings n walls or to support heavy roofs. One of the arch buildings they created was the Arch of Constantine. It is the latest triumphal arch in Rome. It was also the largest setup in Rome since the end of the Severna dynasty. Much of the decorations provided were from earlier monuments of Trojan, Hadrian, and Marcus Tutorials. They included the heads of the earlier emperors. The Pantheon was the dome creation. This was the Temple of all Gods. The dome of the Pantheon is 142 feet high. It provides light entering through a central opening to the sky. Pantheon’s dome is still the world’s largest unrecorded concrete dome to this date. Sanitaria, G. M. (2011, September). Roman architecture. Retrieved from http:// www. Marilynn. Com/ancient Rome/roman architecture. HTML 2. What features did Early Christian architects take from Roman builders, and what new techniques did they develop? Discuss specific features from three buildings. The features that the Early Christian architects took from Roman builders were the arches and domes. But with new techniques they developed such as the flying turrets. For example the Old Saint Peter church resembled Roman basilicas. It had a wide central nave with side aisles and an apse at the end. Unlike Roman basilicas, instead of doorways on one long side opening onto an aisle, they had an obvious long axis opening. When the worshipers arrived in Saint Peters nave, they were able to view the altar in the apse, framed by the chancel arch dividing the nave from the transept. Santa Sabina is another basilicas church. It is famed for its 5th-century wooden doors that have biblical scenes carved in them. It also has 24 columns of marble completely coordinated with Corinthian columns and bases. The spandrels of the arches have decorative marble designs in green and purple to represent the Eucharist. The Santa Stanza Church design was along the favorite of the Western Christian world. One of the most important historical aspects of Santa Stanza were the use of Apses and Central Dome which give prime examples of early Christian art and architecture. Klein, F. S. Gardener’s Art Through the Ages; The Western Perspective. Fourth teen Edition. How to cite Roman and Early Christian Architecture, Papers

The Density of Glass for A Graphical Determination

Question: Disucss about The Density of Glass for A Graphical Determination? Answer: Purpose The main purpose of this experiment is to create a graph which will be used for the determination of the density of the glass i.e. in this case glass marbles. Introduction (A) Density (D) - Density is a physical property of a substance which is generally used to identify that substance. Density (D) may be explained as the ratio of the mass (m) of a substance to the volume (v) occupied by that mass. Density is an intensive property of a matter thus it can be said that the value of density of a matter is independent of the quantity of the matter present. It generally shows how tightly matter is crammed together. The formula for density is given as follows: D= Mass/ Volume or D= m/v The density may be expressed as: g/cm3 or g/mL (1cm3 = 1mL) (B) Indirect Measurement- Indirect Measurement may be defined as a measuring approach to measure things using alternative measurements and properties. Indirect measurement generally involves properties such as Pythagoras theorem, proportions, similar triangles or polygons and others. (C) % Deviation- Percentage deviation may be defined as the measure of how much the mean of a set data differs from the known theoretical value. The formula for the Percentage (%) deviation may be given as follows: Percentage Error = [(Observed value True Value) / True Value] 100 % (D) Dependent variable Dependent Variable may be defined as the variable which is measured in an experiment. The dependent variable generally depends upon the independent variables. For example in this experiment density of glass is the dependent variable. (E) Independent Variable An independent variable may be defined as the variable that can be changed during an experiment. It does not change due to the changes in the other variable. In this example the Mass of water and mass of glass are the examples of independent variable. These variables are generally used to establish some relationship to determine what effect is caused in the dependent variables due to the changes made in the independent variable. (F) Slope of a Line The slope of a line may be defined as the ratio of the amount of by which values on the Y-axis of a graph increases when certain amount is increased in the values in X-axis. Slope of the line how steep is the line, i.e. how much y increases as x increases. The value of the slope is constant everywhere in the line. The slope of a line may be expressed as: Slope = (change in y)/ (change in x) or m = y / x (G) Straight Line Equation Form- It may be described as straight line on the coordinate plane which can be described by the following equation: y = mx + b Where, x, y = Coordinates of any point on the line m= slope of the line b = It is the intercept (where the line crosses the y- axis) The straight line equation form is generally used for two main purposes which are: as a compact way of defining a particular line and to locate all the points on a line. (H) Primary Date - Primary data may be defined as the data which is either observed or collected from the first-hand experience during an experiment or observation. (I) Secondary data Secondary data may be defined as the information that is obtained after interpreting or evaluating the primary data. Diagram Procedure Following is the procedure for conducting the experiment: 1) Put on your safety goggles and Lab apron: 2) Take 100 mL graduated cylinder and fill with about 40mL of tap water. Record this on your primary data table. 3) Make Sure that the above volume is out to the correct precision. 4) Find the combined mass of the graduate cylinder and tap water. Record this on your primary data table. 5) Using your container of glass marbles, take out one marble and place it into the water in the graduate cylinder. Record both the new volume and the new mass into your primary data table. 6) Continue the above process for all the fifteen marbles that are in your container. 7) When completed, pour out the excess water into the drain and place the wet marbles on a paper towel for drying and returning to its container. Primary Data Table Trial Mass of water GC Glass(g) Volume of water Glass marbles (mL) 0 ( Zero Marble) 179.44 g 40.0 mL 1 183.76 g 41.0 mL 2 188.13 g 42.0 mL 3 192.27 g 44.0 mL 4 196.61 g 46.0 mL 5 202.30 g 48.0 mL 6 206.75 g 50.0 mL 7 210 .22 g 52.0 mL 8 214.32 g 54.0 mL 9 218.07 g 56.0 mL 10 222.33 g 58.0 mL 11 226.51 g 60.0 mL 12 230.89 g 61.0 mL 13 235.09 g 62.0 mL 14 239.07 g 63.0 mL 15 243.35 g 64.0 mL Processing the data (Secondary data) 1) For the trial #1 as a sample calculation, determine the mass of the glass in the graduate cylinder. From the Primary data table we can obtain the following data, i) The Mass of water+ Mass of Graduated cylinder+ Mass of Glass marble for Trail #1 = 183.76g-------- (a) ii) The Mass of Water + Mass of Graduated cylinder from Trial #0 = 179.44 g ------- (b) Thus, the in order to obtain the mass of the glass marble for the Trial #1 , subtract (b) from (a). Hence, (183.76 179.44) g = 4.32 g The mass of the glass marble for the Trial #1 is 4.32 g. 2) Place the above mass of glass for trail #1 in the secondary data table. 3) There is now no need to show the method again, but simply place the total glass mass that was in the graduate cylinder for the rest of the 15 trials in the secondary data table. 4) For the trail# 1 as a sample calculation, determine the volume of the glass marble in the graduate cylinder. From the Primary data table we can obtain the following data, i) (Volume of the water + Volume of marble) for Trial #1 = 41.0 mL --------- (a) ii) Volume of water for the trial #0 = 40.0 mL -------- (b) Thus the volume of glass marble in the graduate cylinder = (a) (b) = 1.0 mL. 5) Place the above volume of glass for trial #1 in the secondary data table. 6) Determine the volume of the total glass in the graduate cylinder for each of the remaining trials and place the data in the secondary data table. 7) For trial #1 as a sample calculation, determine the density of the glass marble in the graduate cylinder for each of the remaining trials and place that data in the secondary data table. We know that the Density(D) = Mass/ volume For the trial #1, Mass (m) = 4.32 g Volume of the Glass marble (v) = 1.0 mL Thus, D = m/v = 4.32/ 1.0 = 4.32 g/mL Thus the density of the glass marble for the trial #1 is 4.32 g/mL. 8) Determine the density of the total glass in the graduate cylinder for each of the remaining trials and place that data in the secondary data table. 9) Using your secondary data table and the graphing handout sheet, construct a graph of Mass of glass versus volume of glass versus volume of the glass. Secondary data table Trial # Mass of Glass (g) Volume of Glass (mL) Density of Glass (g/mL) 1 4.32 g 1.0 mL 4.3 (g/mL) 2 8.69 g 2.0 mL 4.3 (g/mL) 3 12.83 g 4.0 mL 3.2 (g/mL) 4 17.17 g 6.0 mL 2.9 (g/mL) 5 22.86 g 8.0 mL 2.9 (g/mL) 6 27.31 g 10.0 mL 2.73 (g/mL) 7 30.78 g 11.0 mL 2.79 (g/mL) 8 34.88 g 13.0 ml 2.68 (g/mL) 9 38.63 g 15.0 mL 2.58 (g/mL) 10 42.89 g 16.0 mL 2.68 (g/mL) 11 47.07 g 18.0 mL 2.62 (g/mL) 12 51.45 g 20.0 mL 2.57 (g/mL) 13 55.65 g 21.0 mL 2.65 (g/mL) 14 59.63 g 23.0 mL 2.59 (g/mL) 15 63.91 g 24.0 ml 2.66 (g/mL) Graph Conclusions and Questions 1) The main purpose of this experiment is to determine the density of the class marble from the Graph of mass vs volume. 2) The range of Density may be calculated from the Difference between the maximum value of density and the minimum value of density. Thus Range of Density = Dmax DMin = (4.3 2.57) g/mL = 1.73 mL 3) The percentage deviation is determined in the following manner, of all the density = 44.15 g/mL Arithmetic mean = 44.15/15 = 2.94 Percent Deviation = [(4.3 -2.94)/2.94] 100 = 46.25% 4) The easiest instrument that was used during the experiment was scale. It was easier to use because there was very little chances of the error in reading in scale and it is easier to take reading from a scale. 5) a) Weight = 0.250 pounds = 113.398gm Density obtained from the graph = 2. 94 g/mL Volume = 113.398/2.94 = 38.57 mL b) Dimensions of glass window are as follows: Length = 32 inch = 81.28 cm Breadth = 22 inch = 55.88 cm Width = 0.125 inch = 0.3175 cm Volume = L B W = (81.28 55.88 0.3175)cm = 1442.06 cm3 1000 cm3 = 1 L Thus, 1442.06 cm3 = 1.442 L = 1442 mL Mass = 1442 2.94 = 4239.48 g