course Mth 271 I had a problem with question 10. I think it is an error in the typing but I am not sure. 單ƖІFassignment #003
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13:30:48 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> 2(ab)+2(ac)+2(bc) = surface area of rectangular solid 3 =a 4=b 6=c (2*3*4)+(2*3*6)+(2*4*6)=108 confidence assessment: 2
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13:31:24 A 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.
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RESPONSE --> I had to look up the equation but now I see why the equation is what it is. That makes a lot of sense. self critique assessment: 3
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13:38:54 `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?
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RESPONSE --> When thinking of a cylinder I think of a can of soup. The label would be the curved sides of the cylinder and if it was layed out flat it becomes a rectangle. One side of the rectangle is the height which is 12 m. So (2 pi * 5) * 12 = the surface area which is 376.9m^2 if it was closed you would add the 376.9m^2 to the ends which would be considered 2 circles so (2 * pi *5^2) 157.1m^2 + 376.9m^2 = 534m^2 confidence assessment: 2
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13:41:10 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.
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RESPONSE --> I understand. Would it be better if I left it as pi m^2 or does it matter if I go ahead and multiply the pi into the answer to make it just ___ m^2?
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13:43:09 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> 4 * pi * 3^2 = 36 pi cm^2 or 113.1 cm^2 confidence assessment: 3
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13:45:53 The 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.
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RESPONSE --> I knew how to do the problem. I misread though and did not read that it actually said DIAMETER intead of RADIUS. I had the right formula but I used the diameter instead of the radius. The problem would have been correct if I did 3/2 because half of the diameter is the radius. self critique assessment: 3
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13:47:33 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> a^2 + b^2 = c^2 25+81 = 106 sqrt 106 = 10.3 m = c = hypotenuse confidence assessment: 3
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13:48:22 The 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.
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RESPONSE --> no self critique needed self critique assessment: 3
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13:49:37 `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?
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RESPONSE --> 6^2 = 4^2 + a^2 36 = 16 + a^2 20 = a^2 sqrt 20 = a= leg =4.5m confidence assessment: 3
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13:49:48 If 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.
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RESPONSE --> no self critique needed self critique assessment: 3
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13:51:24 `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?
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RESPONSE --> D = M/V so d = 700 g / (4*7*12) d = 700 g / 336cm^3 d= 2.08 g / cm^3 confidence assessment: 3
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13:51:38 The 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).
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RESPONSE --> no self critique needed self critique assessment: 3
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14:07:26 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
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RESPONSE --> 3000 kg/ m^3 = M/ ( 4/3*pi*4^3) 3000 kg/m^3 = M / (268.08 m ^3) 3000kg/m^3 * 268.08 m ^3 = M =804,240 kg confidence assessment: 3
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14:08:11 A 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.
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RESPONSE --> I understood this problem. My answer was figured with the pi already multiplied into the answer so the answer is in KG and not pi kg. self critique assessment: 3
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14:09:47 `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?
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RESPONSE --> 4-2/ (6-10) = av density per cm^3 so 4.5 g/cm^3 is the density. confidence assessment: 2
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14:11:29 The 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.
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RESPONSE --> I'm not sure why I used what I did to get the answer that I did before. I was trying to relate it back to the average slope I guess. I see now that I needed to find the mass which is D*V = m then divide it by the total volume to get the average density. self critique assessment: 2
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14:16:04 `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?
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RESPONSE --> average density = total mass / total volume 2100 kg /m^3 x 27 m^3 = 56,700 kg mass of sand 8000 kg/m^3 x 3 m^3 = 24,000 kg 56,700+24000= 80,7000 kg = total mass 27 m^3 = 3 m^3 = 30 m^3 = total volume 80700/30 = av density = 2,690 kg/m^3 confidence assessment: 3
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14:16:25 We 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..
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RESPONSE --> ok self critique assessment: 3
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14:21:46 `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?
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RESPONSE --> 1700000 m^2 x .015 m = 25,500 m^3 860kg/m^3 = m/ 25,500 m^3 m= 21,930,000 kg confidence assessment: 2
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14:24:18 The 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.
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RESPONSE --> I am not sure if I got the right answer or not. I got the same answer up to the volume. This could be a misprint but you then changed the volume to 24,400 instead of 25,500. Either way when I punch it in my calculator I am a digit off. You have 2,193,000 and I got 21,930,000 when multiplying by the volume 25,500. self critique assessment: 1
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14:26:42 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> you can look at it as two parts added together. The top and bottom bases can be looked at as two circles so you can use 2* pi r^2 to find the area of those two circles. then you would add the middle which can be looked at as a label and when you remove a label from the can it is a rectangle. So you can multiply the height by 2* pi r. so A = (2* pi * r^2) + (2* pi* r) * H confidence assessment: 3
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14:27:05 The 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.
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RESPONSE --> ok self critique assessment: 3
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14:27:39 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> surface area of a sphere is 4* pi r*3^2 confidence assessment: 3
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14:27:45 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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RESPONSE --> ok self critique assessment: 3
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14:28:18 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> I know that density is mass divided by volume. I can't say much more than that. confidence assessment: 1
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14:29:44 The 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'
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RESPONSE --> Okay I see that there is a difference when speaking of average density and density unless the object is homogeneous and uniform then density would be the same as average density. self critique assessment: 2
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14:30:24 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> By multiplying the average density by the mass we get the volume. confidence assessment: 3
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14:31:26 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> Okay. Im not sure how I managed to get that confused. I knew what I was saying but instead of writing it down on paper I thought I used the right properties to get it into the form of volume = mass/ av density but I was wrong. I now understand. self critique assessment: 3
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14:32:04 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I wrote down all of the formulas so I can therefore have them at my despense if I do not remember them when they are needed. I sorted out all of the information and circled what I needed to review over. confidence assessment: 3
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