course Phy 231 This ends the third assignment. ëó;Çïcð{jߤ‡ÎÀ®öˆå²ø¯
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16:41:20 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> volume = 3 * 5 * 7 = 105 cm^3 confidence assessment: 3
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16:43:20 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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RESPONSE --> I should remember that V = A * h is the broader formula and should be used instead of V =L*W*h self critique assessment: 3
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16:44:04 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> V = A * h, V = 48 * 2 = 96 m^3 confidence assessment: 3
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16:44:36 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> OK self critique assessment: 3
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16:45:20 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> V = A * h, V = 20 * 40 = 800 m^3 confidence assessment: 3
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16:45:39 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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RESPONSE --> OK self critique assessment: 3
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16:47:36 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> Area = 3.14 * 5^2 = 78.5 cm^2 V = A * h, V = 78.5 * 30 = 2355 cm^3 confidence assessment: 3
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16:48:16 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
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RESPONSE --> OK self critique assessment: 3
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16:54:29 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> I estimate the radius is 3in and height is 5in. Volume = A * h A = 3.14 * 3^2 = 28.26 V = 28.26 * 5 = 141.3 in^3 confidence assessment: 3
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16:54:55 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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RESPONSE --> OK self critique assessment: 3
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17:00:44 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> V = A * h * 1/3 V = 50 * 60 * 1/3 = 1000 cm^3 confidence assessment: 3
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17:00:56 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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RESPONSE --> OK self critique assessment: 3
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17:02:25 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> V = A * h * 1/3 V = 20 * 9 * 1/3 = 60 m^3 confidence assessment: 3
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17:02:40 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
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RESPONSE --> OK self critique assessment: 3
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17:05:39 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> V = 4/3 * 3.14 * r^3 V = 4/3 * 3.14 * 64 = 267.95 m^3 confidence assessment: 3
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17:06:02 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
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RESPONSE --> OK self critique assessment: 3
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17:08:24 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> r = 7000, V = 4/3 * 3.14 * 343,000,000 V = 1,436,026,666.67 km^3 confidence assessment: 3
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17:08:45 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> OK self critique assessment: 3
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17:09:37 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> We take the area of the base and multiply it by altitude or height. V = A * h confidence assessment: 3
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17:09:50 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> OK self critique assessment: 3
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17:10:34 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> We take the area of the base multiply it by the height and then multiply it by 1/3. confidence assessment: 3
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17:10:43 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> OK self critique assessment: 3
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17:11:12 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> V = 4/3 * pi * r^3 confidence assessment: 3
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17:11:20 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> OK self critique assessment: 3
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17:12:02 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have written down the formulas of each geometric object in question. confidence assessment: 3
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Èâºþ—øMä†JåÚ±îí¤ýD÷«qžôˆ“†¬» assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 09-05-2008
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17:55:37 `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 --> SA = 2Lw * 2Lh * 2wh SA = 2*3*4 + 2*4*6 + 2*3*6 = 24 + 48 + 36 = 108 m^2 confidence assessment: 3
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17:55:47 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 --> OK self critique assessment: 3
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18:08:58 `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 of side = 2 * pi * r * h SA of side = 2 * 3.14 * 5 * 12 = 565.2 m^2 total SA = 2pi * r^2 + 2pi * 5 * 12 = 2 * 3.14 * 5^2 + 2 * 3.14 * 5 * 12 = 533.8 m^2 confidence assessment: 3
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18:09:38 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 --> OK confidence assessment: 3
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18:12:12 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> SA = 4 * 3.14 * 1.5^2 because radius is 1/2 of diameter, r = 1.5 SA = 28.26 cm^2 confidence assessment: 3
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18:12:26 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 --> OK self critique assessment: 3
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18:16:45 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> c^2 = a^2 + b^2 c^2 = 25 + 81 = 106 c = 10.30 confidence assessment: 3
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18:17:09 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 forgot to put in units. answer: 10.3m self critique assessment: 3
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18:18:14 `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 --> c^2 = a^2 + b^2 36 = 16 + b^2 b^2 = 20, b = 4.47m confidence assessment: 3
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18:18:22 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 --> OK self critique assessment: 3
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18:22:03 `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 Density = 700g/(4 * 7 * 12) = 2.08 g/cm^3 confidence assessment: 3
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18:22:27 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 --> OK self critique assessment: 3
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18:27:06 `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 --> Density = mass/volume volume = 4/3 * 3.14 * 4^3 = 267.95 m^3 mass = density * volume mass = 3000 kg/m^3 * 267.95 m^3 mass = 803,850 kg confidence assessment: 3
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18:27:48 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 --> OK self critique assessment: 3
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18:33:56 `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 --> Density = mass/volume mass = density * volume mass1 = 4 g/cm^3 * 6 cm^3 = 24g mass2 = 2 g/cm^3 * 10 cm^3 = 20g total mass = 24 + 20 = 44g total volume = 6 cm^3 + 10 cm^3 = 16 cm^3 Density = 44g / 16 cm^3 = 2.75 g/cm^3 confidence assessment: 3
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18:34:06 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 --> OK self critique assessment: 3
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18:49:29 `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 --> Density = mass/volume total volume = 27 m^3 + 3 m^3 = 30 m^3 mass1 = 2100 kg/m^3 * 27 m^3 = 56700 kg mass2 = 8000 kg/m^3 *3 m^3 = 24000 kg total mass = 56700 + 24000 = 80700kg density = 80700/30 = 2690 kg/m^3 confidence assessment: 3
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18:50:24 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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18:56:36 `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 --> density = mass/volume mass = density * volume volume = 1,700,000 m^2 * .015m = 25,500 m^3 mass = 860 kg/m^3 * 25,500 m^3 = 21,930,000 kg confidence assessment: 3
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18:56:49 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 --> OK self critique assessment: 3
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18:58:11 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> SA = (2 * pi * r^2) + (2 * pi * r * h) confidence assessment: 3
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18:58:19 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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18:59:07 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> SA = 4 * pi * r^2 confidence assessment: 3
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18:59:15 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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19:00:23 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is to measure how tight matter is packed in to the object. Density = mass/volume confidence assessment: 3
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19:00:33 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 --> OK self critique assessment: 3
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19:01:07 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Density = mass/volume volume = mass/density confidence assessment: 3
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19:01:17 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> OK self critique assessment: 3
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19:01:55 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have taken notes of all formulas for geometric shapes used. confidence assessment: 3
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