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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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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Cell Biology- Osmosis‚ Cell Size and Diffusion and Enzymes 1.0 INTRODUCTION Cells are the basic building blocks of all living things. They provide structure for the body‚ take in nutrients from the food‚ convert those nutrients into energy‚ and carry out specialized functions. Cells also contain the body’s genetic material and can make copies of themselves. A cell is also a metabolic compartment where many different chemical reaction occur. There are two types of cells‚ eukaryotic and prokaryotic
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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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Investigating Ratios of Areas and Volumes In this portfolio‚ I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b‚ such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures‚ several different values for n will
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results Surface area to volume ratio Time taken for HCL acid to diffuse (seconds) 14 38 9 90 7.3 139 6.5 178 6 185 3. Do your results support your hypothesis‚ use date from your table and graph to support your answer‚ you should also identify any anomalous results. My hypothesis was ‘as the surface-area-to-volume ratio decreases the diffusion rate increases’ my results support my hypothesis‚ because in my table of results when the surface area to volume ratio is at 14 the diffusion time is
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Risk 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
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Surface area Surface area is the measure of how much exposed area a solid object has‚ expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces‚ such as a sphere‚ are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods
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INTERNAL ASSESMENT DATA COLLECTION AND PROCESSING CONCLUSION AND EVALUATION RELATIONSHIP BETWEEN SURFACE AREA AND THE VOLUME OF A CELL‚ HOW IT AFFECTS THE RATE OF DIFFUSION Nazirul Ibrahim 04.10.2011 DATA COLLECTION AND PROCESSING We have determined the following variables: * dV- the time for the color indicator phenolphthalein to reach the center of the cube (diffusion) * iV- the size of the cube * cV- the amount of Sodium Hydroxide the perfect dimensions of the cube
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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.
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