course Phy 201 Hello. Have a great day.
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Given Solution: `aA rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area? ********************************************* Your solution: SA = 2pi x r x h. 2pi x 5m x 12m = 376.8m^2 To find the surface area plus the top and the bottom, I would add the area of each circle. A = pi x r^2. Pi x (5m)^2 = 78.5m^2. 78.5m^2 x 2 = 157 m^2 376.8m^2 + 157m^2 = 533.8m^2 Confidence Assessment: 2
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Given Solution: The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q003. What is surface area of a sphere of diameter three cm? ********************************************* Your solution: SA = 4pi x r^2. Half the diameter of 3 cm is 1.5 cm. 4pi x (1.5cm)^2 = 28.26 cm^2 Confidence Assessment: 3
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Given Solution: `aThe surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2. NOTE TO STUDENT: While your work on most problems has been good, you left this problem blank and didn't self-critique. You should self-critique here. • For example you should acknowledge having made note of the formula for the surface area of the sphere, which I expect you didn't know before. I expect from your previous answers that you are very capable of applying the formula once you have it, and based on this history you probably wouldn't need to self-critique that aspect of the process. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters? ********************************************* Your solution: I remember having to find hypotenuses in geometry, but that was all the way back in 9th grade! I think the formula is a^2 + b^2 = c^2. So (5m)^2 + (9m)^2 = 106m^2. This does not exaclt look right to me, but I cannot remember anything else. Confidence Assessment: 1
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Given Solution: `aThe Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the formula correct! I just didn’t take the square root of the answer, which would then have given me the right solution. Self-critique Rating: 3 ********************************************* Question: `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg? ********************************************* Your solution: C^2 = a^2 + b^2. C^2 – a^2 = b^2. B = sqrt(c^2) – sqrt(a^2). Sqrt(6m^2) – sqrt(4m^2) = 2m. Confidence Assessment: 2
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Given Solution: `aIf c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The only mistake I made was putting the sqrt in front of each c and a. now I know that the answer is the sqrt of (c-a). Self-critique Rating: 3 ********************************************* Question: `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm? ********************************************* Your solution: Density = mass/volume. First I need to find the volume of the rectangular solid. V = l x w x h. V = 4cm x 7cm x 12cm = 336cm^3. Now I can find the density. D = 700g/336cm^3 = 2.08g/cm^3. Confidence Assessment: 3
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Given Solution: `aThe volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3). NOTE TO STUDENT: You did not respond to the question and did not self-critique. You would be expected to address the question, stating what you do and do not understand. • For example you should understand what a rectangular solid with dimensions 4 cm by 7 cm by 12 cm is, and how to find its volume and surface area. You might not know what to do with this information (for example you might well not understand that it's the volume and not the surface area that's related to density), but from previous work you should understand this much, and should at least mention something along the lines of 'well, I do know that I can find the volume and/or surface area of that solid' in a partial solution. • The word 'density' is clearly very important. Even if you don't know what density is, you could note from the statement of the problem that its units here are said to be 'grams per cubic centimeter'. Having noted these things, you will be much better prepared to understand the information in the given solution. I expect you understand the first statement from previous knowledge (you should have this understanding from prerequisite courses, and if not you encountered it in the preceding 'volumes' exercise): 'The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.' It would of course be appropriate to ask a question here if necessary. It is likely that, as is the case with many students, the concept of density is not that familiar to you. However if this wasn't addressed specifically in prerequisite courses, those courses would be expected to prepare you to understand this concept. The statement 'Its density in grams per cm^3 is the number of grams in each cm^3.' serves as a definition of density. In your self-critique you should have addressed what what this phrase means to you, and what you do or do not understand about it The next phrase is 'We find this quantity by dividing the number of grams by the number of cm^3.' You would be expected to understand that this phrase is related to the preceding, and as best you can to address the connection. At this point many students would need to ask a question, and it would be perfectly appropriate to do so (or to have done so regarding previous statements). The subsequent phrase 'density = 700 grams / (336 cm^3) = 2.06 grams / cm^3' is an illustration of the ideas and definitions in the preceding statements. A reasonable self-critique would demonstrate your attempt to understand this statement and its connection to the preceding. Once again questions would also be appropriate and welcome. The remainder of the given solution provides additional information. Your prerequisite courses will ideally have prepared you to understand these ideas, and for students entering the Mth 173-4 sequence or the Phy 201-202 or 231-232 sequence you should be able to either completely understand all the details of the given solution, or address them in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter? ********************************************* Your solution: D = m/v. We can find the volume of the sphere by using the formula V = 4/3pi x r^3. V = 4/3pi x (4m)^3 = 268m^3. Now if we rearrange the density formula so that we are finding mass, m = D x v. m = 3000kg/m^3 x 268m^3 = 804000kg. Confidence Assessment: 3
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Given Solution: `aA average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object? ********************************************* Your solution: I’m not sure how we would use the information about the volume of each of the materials. If we use only the densities given then the average density of the object would be (4g/cm^3 + 2g/cm^3)/2 = 3g/cm^3. Confidence Assessment: 2
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Given Solution: `aThe first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand now that we had to find the mass of each material. We wanted to find the average density of the object and that would of course be total mass/total volume = average density. Self-critique Rating: 3 ********************************************* Question: `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box? ********************************************* Your solution: The box has a volume of 2m x 3m x 5m = 30m^3. The total density of the sand is 27m^3 x 2100 kg/m^3= 56700kg/m^3. The mass then of the sand is 56700kg/m^3 x 30m^3 = 1701000kg. The total density of the cannon balls is 3m^3 x 8000kg/m^3 = 24000kg/m^3. The mass of the cannonballs is therefore 24000kg/m^3 x 30m^3 = 720000kg. Adding the mass of the sand and the mass of the cannonballs = 2421000kg. Average density = total mass/total volume. Average density = 2421000kg/30m^3 = 80700kg/m^3. Confidence Assessment: 1
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Given Solution: `aWe find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I misread the question. I did not understand from the wording that the 27m^3 of sand and the 3m^3 of cannon balls was the actual volume. Self-critique Rating: 3 ********************************************* Question: `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick? ********************************************* Your solution: I think the first part of the question is asking for volume. So V = A x h. V = 1700000m^2 x .015m = 25500m^3. The mass is then m = D x v. m = 860kg/m^3 x 25500m^3 = 21930000kg. Confidence Assessment: 2
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Given Solution: `aThe volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder? ********************************************* Your solution: Surface area of a cylinder = (2pi x r x h) + (2pi x r^2) where r is the radius and h is the height or altitude. Confidence Assessment: 3
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Given Solution: `aThe curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere? ********************************************* Your solution: Surface area of a sphere = 4pi x r^2. Confidence Assessment: 3
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Given Solution: `aThe surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'. ********************************************* Your solution: If the density of an abject is 3g/cm^3, that means that each cubic centimeter has a mass of 3g. Confidence Assessment: 2
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Given Solution: `aThe average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density' &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): D = m/v Self-critique Rating: 3 ********************************************* Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume? ********************************************* Your solution: If we know the average density of an object and we know the mass of the object, we would simply rearrange the formula to find volume. This would be v = m/D. Confidence Assessment: 3
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Given Solution: `aSince mass = ave density * volume, it follows by simple algebra that volume = mass / ave density. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q015. Part 1 Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. These principles are a bit more difficult than previous principles. But if I remember how to find volume, then I can find density and mass. I need to remember though that the average density of an object is = total mass/total volume.