course Phy 201 I'm I taking too long to do my work? I feel like I can't get much accomplished in a day and that I'm getting farther and farther behind. JϗlІ}sassignment #003
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15:15:05 `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 --> I'm not sure of my answer, but I think because the volume of a rectangular solid is V = L * w * h then the surface area would be: 2L * 2w * 2h. So, surface area = 2(3m) * 2(4m) *2(6m) = 576 m^3. confidence assessment: 1
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15:27:57 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 knew there was two faces for each but I didn't know mathmatically how to express the problem. Can it also be written as ( w * h ) + ( L * h ) + ( L * w )? self critique assessment: 2
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15:42:19 `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 --> The formula for the surface area of a cylinder is: A = 2 pi * r * L + 2 pi * r^2 To obtain the surface area of the curved sides of a cylinder the formula will be: A = 2 pi * r * L. The rest of the first formula is not need because it finds the surface area of the circles at each end. So the answers will be: A = 2 pi * 5m * 12m = 120 pi m^2 ( curved sides) A = ( 2 pi * 5m * 12m ) + ( 2 pi * ( 5m)^2 ) A = 120 pi m^2 + 50 pi m^2 A = 170 pi m^2 (the entire surface area of a cylinder) confidence assessment: 2
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15:52:17 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 feel my response was appropriate. confidence assessment: 3
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16:02:24 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> The formula for surface area of a sphere is: A = 4 pi r^2. A = 4 pi * ( 3cm )^2 A = 4 pi * 9 cm^2 A = 36 pi cm^2 You can take the area of a circle and multiply it times 4. confidence assessment: 3
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16:07:35 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 left the diameter of 3 as the radius. I didn't read the question correctly. I know where you got the radius, divide the diameter by 2. self critique assessment: 2
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16:20:00 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> Here you would use the Pythagorean Theorem. ( leg 1 )^2 + ( leg 2 )^2 = ( hypotenuse ) ^2. I will use a^2 + b^2 = c^2 where a = 5 meters and b = 9 meters. ( 5m )^2 + ( 9m )^2 = c^2 106 m^2 = c^2 sq. root of 106 m^2 = c Thus the hypotenuse is equal to the square root of 106 m^2. confidence assessment: 3
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16:23:45 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 --> I left my answer in square root form. I should not have done this. I should have finished the problem to get my answer in the correct units. self critique assessment: 2
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16:33:13 `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 --> Use the Pythagorean Theorem. The length of the leg in question will be b^2. ( 4m ^2 ) + b^2 = ( 6m^2 ) 16 m^2 + b^2 = 36 m^2 b^2 = 20 m^2 b = sq. root of 20 m^2 b = 4.5 meters 4.5 meters is the length of the other leg. confidence assessment: 3
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16:34:52 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 --> My answer on the calculator was 4.47.... Should have not rounded the 4 up to 5 ? self critique assessment: 2
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16:47:49 `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 --> The formula for density is mass / volume. The volume is not given in this problem so we must find it first before we can complete the problem. V = L * w * h or V = A * h V = 4cm * 7cm * 12 cm V = 336 cm^3 Now we can find the density. Density = 700 g / 336 cm^3 Density = 2.08 g / cm^3 confidence assessment: 3
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16:54:54 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 --> Notes taken self critique assessment: 3
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17:09:54 `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 solve for the mass of a sphere, using the formula for density, the volume of the sphere must be calculated. V = 4/3 pi * r^3 V = 4/3 pi * ( 4m )^3 V = 4/3 pi * 64 m^3 V = 256/3 pi m^3 V = 268 m^3 Now we can use the density formula to solve for mass. 3000 kg / m^3 = M / 268 m^3 ( 268 m^3 ) * 3000 kg / m^3 = ( M / 268 m^3 ) * 268 m^3 The 268 m^3 on the right side both cancel leaving: ( 268 m^3 ) * 3000 kg / m^3 = M M = 80,400 kg The mass of the sphere is 80,400 kg. confidence assessment: 3
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17:18:49 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 --> In the answer I gave I think I didn't put enough zeros. The answer should have been 804,000 kg. I should have waited to round my answers until the end. If I would have waited to round my last answer it would have been: 256,000 pi kg = 804,247 kg I think it would have been a more accurate answer. self critique assessment: 2
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17:28:33 `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 --> To get the average density, add the two given densities and divide by 2. 4 g / cm^3 + 2 g / cm^3 = 6 g / cm^3 ( 6 g / cm^3 ) / 2 3 g / cm^3 is the average density. confidence assessment: 3
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17:39:27 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 was going to find the mass of each one, but thought I was reading to much into the question, so I answered it the way I did. So I should have realized those other numbers weren't just in the problem for entertainment...I guess it would be entertainment, to see how many people answered it the same way I did. I understand the problem, but I did take notes and answered the problem correctly in my notebook. self critique assessment: 2
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18:02:38 `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 solve this problem find the mass of both the sand and the cannonballs. mass = density * volume mass (sand) = 2100 kg / m^3 * 27 m^3 mass (sand) = 56,700 kg mass(cannonballs) = 8000 kg / m^3 * 3 m^3 mass(cannonballs) = 24,000 kg To find the average density take the total mass and divide it by the total volume: ( 56,700 kg + 24,000 kg ) / ( 27 m^3 + 3 m^3 ) 80,700 kg / 30 m^3 Average density = 2690 kg /m^3 ( You can say that the box is filled to the top and level because you can find the volume of the box by multiplying 2m * 3m * 5m = 30m^3, which is equal to the total volume of the sand and cannonballs) confidence assessment: 3
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18:03:44 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 --> I learn from the previous question so I could apply it to this question. self critique assessment: 2
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18:20:55 `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 --> "" How many cubic meters of oil are there..."" can be considered as finding the volume of the oil. 1,700,000 m^2 is the 'area' of the oil slick. To find the volume take the area and multipy by the depth of the oil: V = A * h V = 1,700,000 m^2 * 0.15 m V = 255,000 m^3 There are 255,000cm^3 of oil in the oil slick. To find the mass multiply density by volume. Mass = 255,000 m^3 * 860 kg / m^3 Mass = 219,300,000 kg The mass of the oil slick is 219,300,000 kg. confidence assessment: 3
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18:24:07 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 wrote down the thickness as 0.15 meters you put .015 meters. I thought I wrote down what was in the question. Please correct me if I'm wrong on the thickness. I did it correctly except for probably writing down the wrong thickness. self critique assessment: 2
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18:28:14 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> The surface area of a cylinder whose ends are closed is: A = ( 2 pi * r * L ) + ( 2 pi * r^2 ), where 2 pi * r * L is the area on the curved sides and 2 pi r^2 is the area of the circles on each end of the cylinder. confidence assessment: 3
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18:28:42 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 --> self critique assessment: 3
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18:29:55 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The surface area of a sphere is: A = 4 pi r^2, where r is the radius. confidence assessment: 3
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18:30:05 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 --> self critique assessment: 3
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18:33:52 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is defined as mass per unit of volume. Example: you have a substance whose density is 3.64 kg / m^3 means that every cubic meter has a mass of 3.64 kg. confidence assessment: 3
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18:34:09 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 --> self critique assessment: 3
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18:36:53 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> To find volume by using the density formula, density = mass / volume, divide mass into both sides getting the formula V = density / mass. confidence assessment: 3
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18:39:15 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> I got myself confused on the division in the original density formula, I should have used to mass formula to get my answer. self critique assessment: 2
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18:41:40 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I organized my knowledge of the principles by taking notes on the various exercises to use if needed for further reference. confidence assessment: 3
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