course Phy 201 `questionNumber 30000 `q001. There are 10 questions and 5 summary questions in this assignment.
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RESPONSE --> L=Length W=width and H=Height SA=2LH+2HW+2LW =2(6)(4)+2(4)(3)+2(6)(3) =48+24+36 =108m confidence assessment: 2
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11:57:27 `questionNumber 30000 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 --> Uh oh It is meter^2. I didn't put that. Otherwise, I got it. self critique assessment: 2
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12:09:55 `questionNumber 30000 `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 --> SA=(2 pi radius x height) + (2 pi radius^2) For only curved sides, we only use the first part of the equation. So, 2xpix(5)x(12) = 377m^2 For the entire surface area, we use the entire equation: We know the first part = 377 So 377 + 2pi(25)= 377+157=534 m^2 confidence assessment: 2
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12:10:34 `questionNumber 30000 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 --> Okay, I got the same area once pi is actually multiplied through. confidence assessment: 3
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12:12:40 `questionNumber 30000 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> SA=4 pi radius^2 =4pi (1.5)^2 =4 pi (2.25) =9 pi confidence assessment: 3
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12:12:52 `questionNumber 30000 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 understand. self critique assessment: 3
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12:17:33 `questionNumber 30000 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> For this, we use the pythagoream theorem. x^2 + y^2 = r^2 (5)^2 + (9)^2 = r^2 25+81 = r^2 10.3 = r 10.3 is the length of the hypotenuse confidence assessment: 2
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12:17:46 `questionNumber 30000 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 understand. self critique assessment: 3
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12:19:43 `questionNumber 30000 `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 --> 4^2 + x^2 = 6^2 16 + x^2 = 36 x^2 = 20 x=4.5 (approx) 4.5 is the length of the ""other"" side confidence assessment: 3
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12:20:04 `questionNumber 30000 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 --> I understand. self critique assessment: 3
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12:26:40 `questionNumber 30000 `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 --> Density = mass/volume First we have to figure out volume V=area x height =4x7x12=336cm^3 D=m/v =700g/336cm^3=2.08g/cm^3 confidence assessment: 3
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12:27:15 `questionNumber 30000 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 --> I understand. self critique assessment: 3
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12:33:20 `questionNumber 30000 `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 --> First, we have to find the volume of a sphere formula, which is 4/3 pi r^3. So, the Volume is 4/3 pi x 64 = 268.1m^3 Now we use D=m/v the mass is unknown, but we know D is 3000 and V is 268.1 after multiplying 3000 by 268.1, we get 804,300, which is the mass that we were intially solving for. confidence assessment: 3
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12:33:51 `questionNumber 30000 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 --> After multiplying through by pi, my answer was the same. I understand this. self critique assessment: 3
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15:00:48 05-14-2007 15:00:48 `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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NOTES -------> I honestly do not know how to do this. I am all ears on how...
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15:00:56 `questionNumber 30000 `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 --> confidence assessment:
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15:04:55 `questionNumber 30000 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 --> Okay, first one must find the total grams for each material by multiplying the volume by the density. Then, add them together. This number is the total mass that will be divided by the total volume. The volume should be found by merely combining the volumes from material one and two. Divide the total mass (from step one) by the total volume (in step two), and this gives the total density of the two materials. After reading the explanation this makes sense, and I feel like I can apply it again. self critique assessment: 2
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15:11:30 `questionNumber 30000 `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 --> First, one must find the mass of each material by multiplying the density of the material by the given volume. For the first material, the mass is 56,700 kg. For the second, it is 24,000 kg. Then, we find the combined volume which is given twice in the equation--first by the dimensions of the box and second by adding the volumes specifically given for each material. We add the masses of both materials (56,700 +24,000) and get 80,700. Then, we divide this number by the total volume of 30m^2, yielding the total density of 2,690 kg/m^3. confidence assessment: 2
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15:12:36 `questionNumber 30000 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 gave a more specific answer of 2,690, which is still correct. I understand it this time! Yay! self critique assessment: 3
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15:17:29 `questionNumber 30000 `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 --> First, one must find the volume by multiplying 1,700,000x1,700,000x.015=4.335x10^10 Then, since we know the final density and the volume, we merely multiply the final density by the volume to give us the mass: 860 kg/m^3 x 4.335 x 10^10 = 3.7281 x 10^13 confidence assessment: 2
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15:19:50 `questionNumber 30000 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 see that I multiplied 1,700,000 too many times, messing the equation up from the beginning. I didn't realize (although the squared on the m should have been THE clue) that the figure above already represented the area. The rest of the equation I understand...the numbers were merely messed up from me trying to find the volume and multiplying 1,700,000 too many times. self critique assessment: 2
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15:21:05 `questionNumber 30000 `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 can be found by using the formula (2 pi rh) + (2 pi r^2), where the first part of the equation represents the curved sides and the latter part represents the top and bottom of the cylinder. confidence assessment: 3
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15:21:14 `questionNumber 30000 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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15:21:38 `questionNumber 30000 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The formula is SA=4pi(r)^2 confidence assessment: 3
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15:21:48 `questionNumber 30000 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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15:22:41 `questionNumber 30000 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density involves how much mass there is in a certain amount of space, measured by the term volume. Thus, the formula for density incorporates these ideas with the mass being divided by volume. confidence assessment: 3
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15:22:53 `questionNumber 30000 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 understand. self critique assessment: 3
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15:24:10 `questionNumber 30000 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> One can divide the mass by the density D=m/v Dv=M V=m/D confidence assessment: 3
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15:24:26 `questionNumber 30000 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> I understand how this works. self critique assessment: 3
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15:25:25 `questionNumber 30000 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> Like before, I have taken notes thoroughly and have drawn diagrams that supplement my notes. I have documented every example and have explained in writing for myself the areas of the problems with which I had trouble, so that hopefully I will be able to reference this in the future and will not make the same mistakes again. confidence assessment: 3
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