course Mth173 Please remember that I do not have an active email account so send any responses to *******
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16:41:59 `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 --> To get the are of a rectangular solid you would use the formula L*W*H Therefore you would have the following: A=3*4*6, which would give you 72m^3 confidence assessment: 3
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16:43:03 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 --> In evaluating my answer I used the formula for a regular rectangle instead of a rectangular solid. self critique assessment: 3
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16:55:39 `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 --> In computing the answer for the curved sides of the cylinder I used the formula for a the circumference of a circle to find the length of the opened cylinder. C=2*pi*5m = 31.4m. Then I used the formula for a retangle A=L*W to get 31.4m*12m=376.8m^2 to get the surface area of the curved sides of the cylinder. Then I used the volume of a cylinder, V=pi*5^2*12m to get 942 m^2 for the surface area of the cylinder confidence assessment: 0
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17:20:24 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 --> Does it matter if I use 3.14 instead of the word pi to answer the problems? I used the wrong formula for the area of the cylinder. I am totally lost on the answer for the area of the cylinder closed. If I take the area of the two circles, 157m and add this to the area of the sides, 376.8m, I would get 690.8m^2.
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17:23:43 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> For this problem I would use the formula A=4*pi*r^2. You would first find the radiua since the problem is giving the diameter you would divide 3 by 2 to get 1.5 for the radius. A=4*pi*1.5^2 A=9 pi cm^2 confidence assessment: 3
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17:23:59 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 --> Understood self critique assessment: 3
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17:28:40 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> In solving this problem you would use the formula Hypot. = adjac./oppos., which would give you 9m/5m=1.8m confidence assessment: 3
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17:32:49 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 --> Of course I used the wrong formula. We had just did this unit around a month ago and thie formula that I used was what I taught. self critique assessment: 0
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17:39:26 `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 --> If I use the formula A^2+B^2=C^2 then I would get 4^2 + b^2 = 6^2, which then would be, 16 + b^2 = 36, you would then move the 16 to the other side and get 52 and the square root of 52 is 7.2m confidence assessment: 1
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18:13:43 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 --> Unfortunately when I transferred my 16 to the other side of the formula I added it instead of subtracting, which gave me the wrong answer. self critique assessment: 3
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18:31:15 `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 --> In solving this problem I used the formula for the volume of a rectangle and got 7*12*4=336cm^3. You would then divide the 336 cubic centimeters by 700 grams to get .48 grams per cubic centimeter confidence assessment: 3
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18:35:09 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 --> In solving this problem I solved for cubic centimeters per grams instead of grams per cubic centimeters. self critique assessment: 3
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18:41:00 `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 --> To answer this problem you would use the formula for the volume of a sphere. V=4/3 pi r^3, therefore, V=4/3 pi 4^3, V=258/3 pi m^3. You would then take 3,000 kg and divide it by 258/3 pi m^3 to get an average of 34.88 pi kg per cubic meter confidence assessment: 3
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18:44:08 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 read the problem wrong for one thing and read that it had an average weight of 3000 kg. I did not read the fact that it was per cubic meter. I also did some wrong multiplication. self critique assessment: 3
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18:56:20 `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 --> In order to find the average density of this object you would have to multiply the volume of each piece of material to the density of each piece of material to find out the total density for each piece of material. 6cm^3 * 4 g = 24 g and 10 cm^3 * 2 g= 20 g, then you would add the two answers together 20 g + 24 g to get and average density of 44 g confidence assessment: 1
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18:58:00 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 --> Again I needed to go one further step to find the average density. I only did the total mass and forgot to do the average mass. self critique assessment: 2
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19:13:16 `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 --> To find the average density of this problem you would use the volume formula V=2*3*5 therefore, V=30 m^3 for the first part of the question. However, you only used 27 m^3 to put sand in, so you would only multiply 27 * 2100 to get total mass of 56,700 kg/m^3 and for the other three cubic centimeters you would multiply 3 * 8,000 to get 24,000 kg/m ^3. Then to get the average density of the material in the box you would take the total mass (56,700 + 24,000=80,700) and divide by total density (30 m^3 to get an average density of 2,690 kg/m^3 confidence assessment: 3
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19:17:35 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 --> The only thing that I did not do in my problem was round. self critique assessment: 3
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19:29:24 `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 --> The problem tells me that the oil slick covers an area of 1,700,00 m^2 and that the depth is .015 m. To find the volume you would multiply 1,700,00m^2 * .015m = 25,500 m^3 Then to find the mass you would by multiplying the volume (25,500 m^3) to the total density (860 kg/cubic meter and get 21,830,000 kg/m^3 confidence assessment: 0
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19:30:20 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 guess I understood more than I thought. However, I put too may 0's in my answer. self critique assessment: 3
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19:32:44 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> Use the circumference as the length of the sides and multiply this to the height of the cylinder to find the area of the sides of the cylinder. You would then find the area of the circle and double that answer since you have a top and a bottom to a cylinder. You would get your final anser by adding all of your area figures up to get your final answer. confidence assessment: 3
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19:33:09 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 --> 3 self critique assessment:
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19:33:59 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> A = 4 pi r^2 confidence assessment: 3
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19:34:20 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 --> Understood self critique assessment: 3
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19:35:15 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is how full or how much is in an area confidence assessment: 3
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19:36:06 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 --> I defined density and did not put how you would calculate density. self critique assessment: 3
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19:39:08 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> You could divide mass by average density to get volume. confidence assessment: 3
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19:39:25 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> great self critique assessment: 3
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19:40:24 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> Drew diagrams of each problem to be able to visualize the problem and worked each problem out on paper to have the notes for each problem. confidence assessment: 3
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