diffusion. Diffusion is the process that cells use to obtain oxygen‚ water and food. Also‚ how they lose waste substances‚ for example‚ urea and carbon dioxide. Basically‚ Diffusion is when particles move from an area of high concentration to an area of low concentration. The surface area to volume ratio of the cell is an important factor in diffusion. It is the effect of this factor that will be investigated in this practical. Materials used in the practical consist of blocks of agar jelly containing
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Mathematics Volume of Solids Formulae for Volume of Solids Cube | Cuboid | Triangular Prism | Cylinder | Cone | Pyramid | Sphere | AnyPrism | s3 | lwh | ½bhl | Πr2h | 1/3πr2h | 1/3Ah | 4/3πr3 | Ah | A = area of the base of the figure s = length of a side of the figure l = length of the figure w = width of the figure h = height of the figure π = 22/7 or 3.14 1. Compute the volume of a cube with side 7cm. Volume of cube: s3 s = 7cm s3 = (7cm x 7cm x 7cm) = 343cm3 2. Compute
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be found in multicellular organisms. The unicellular and multicellular organisms are linked to cell size and surface area to volume ratio. The experiment for cell size and diffusion was set to see how and how much water can go to the cells. This movement of water is called Osmosis. Osmosis is the movement of water molecules from an area of low concentration (lots of water) to an area of high concentration (little water) through a semi permeable membrane‚ demonstrated in ‘figure 1’. A semi permeable
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AREA (i) The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonal of the rhombus is 22 cm‚ find the length of the other diagonal. (ii) The floor of a rectangular hall has a perimeter 250m. If the cost of paining the four walls at the rate of Rs 10 per m2 is Rs 1500. Find the height of the hall. (iii) A room is half as long again as it is broad. The cost of carpeting the room at Rs
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whatever the temperature may be. The larger the surface area‚ means there can be more “paths” from the sides of the body that are capable of releasing this heat particles‚ and reaching thermal equilibrium faster. This is what happens when a hotter body is subjected to a colder one. Research Question: How does the surface area to volume ratio affect heat loss in organisms? Hypothesis: I hypothesize that the larger the surface area to volume ratio‚ the more heat will be lost and vice versa. In this
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In Parts A and C‚ the relationship between surface area and volume was investigated. Plasticine was formed into a cube and a sphere; both shapes were cut in half. It was found that plasticine volume should not vary‚ two halves have a greater surface area than a whole‚ and cubes have a greater surface area than spheres of the same volume. In Part B‚ the relationship between diffusion and surface area to volume ratio was investigated. Three agar-phenolphthalein-sodium hydroxide cubes of different sizes
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relationship between surface area : volume ratio and heat loss. INTRODUCTION: The aim of this experiment is to investigate and find the relationship between heat loss (of water) and surface area to volume ratio of animals. To investigate this‚ we are going to use three flasks of different volume (as the equivalent the animals) and thus different surface areas filled with water. BACKGROUND: Surface Area : Volume Ratios We will be using the following formula for calculating SA:Vol ratios: SA : Vol Vol :
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Assessment 5 Method 6 Results 6 Analysis/Discussion 7 Conclusion 8 Acknowledgments/Bibliography 8 Table of Contents Background Research As an object becomes bigger its surface area and volume increases but the surface area to volume ratio decreases‚ this is because volume increases quicker than the surface area; as volume is three dimensional. This concept applies to cells and reaction rate because cells need to absorb their food‚ water and oxygen through their cell membrane and depending on the
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Subject: Ratio Analysis of Volume Book Distributors Ltd for the Years 2010 And 2011 A. INTRODUCTION The report is based on the company Volume Book Distributors Ltd. Ratio analysis was calculated for Volume Book Distributors Ltd for both the years 2010 and 2011. The ratios comprised of profitability‚ asset utilization and financial stability ratios. The first question consists of Part A‚ which has the Ratio Calculations‚ and Part B which has the Analysis and Interpretation. B. LIMITATIONS
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average rate of conductivity and the surface area: volume ratio. As the surface area of the Agar cube increases‚ the rate of conductivity will steadily increase‚ too. The trend line of the graph shows as an exponential graph because at one point there will be no more conductivity due to too small or too big Agar cubes. A difference can be seen in the average rate of conductivity depending on surface area because the Agar cubes with a larger total surface area are able to diffuse the ions faster. When
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