course 271
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Given Solution: `aIf 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: ********************************************* Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters? ********************************************* Your solution: Base * height (altitude) = 48 m^2 * 2 m = 96 m^3 Confidence Assessment: 3
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Given Solution: `aUsing 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters? ********************************************* Your solution: Volume = Base * altitude = 20 m^2 * 40m = 800 m^3 Confidence Assessment: 3
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Given Solution: `aV = 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating:3 ********************************************* Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm? ********************************************* Your solution: Base = pi*r^2 = pi*5cm^2 = pi*25cm^2 Volume = Base * Altitude = pi*25cm^2 * 30cm = pi*750cm^3 ≈2,355cm^3 Confidence Assessment: 3
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating:3 ********************************************* Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates? ********************************************* Your solution: Can containing food is small (14 oz) can of popped popcorn. If by volume, this question refers to internal capacity, the internal measurements should be used, in this case: Height (aka altitude, measured from the bottom of the inside of the can to the lip) = 15 cm Diameter of base (measured as maximum measurement across bottom inside rim to inside rim) = 12.5 cm Radius = .5 * diameter = .5*12.5 cm = 6.25cm Area of base = pi*r^2 = pi*(6.25cm)^2 = pi*39.0625cm^2 Volume = base * height = pi*39.0625cm^2*15cm = pi*585.9375cm^3 ≈1,839.84375cm^3 ≈1840 cm^3 If by volume, this question refers to the external volume, as in the minimum size package required to contain it, the external measurements must be used, in this case: Height (aka altitude, measured from bottom of lower rim to top of cap) = 15.5 cm Diameter of base (measures as maximum measurement across rolled rim of top) = 13 cm Radius = .5*diameter = .5*13cm = 6.5cm Area of Base = pi*r^2 = pi*(6.5cm)^2 = pi*42.25cm^2 Volume = Base * Height = pi*42.25cm^2*15.5cm=pi*654.875cm^3≈2,056.3075cm^3 ≈2056cm^3 Confidence Assessment: 3
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Given Solution: `aPeople 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: na ********************************************* Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm? ********************************************* Your solution: I know there is a relationship between the area of a pyramid, and the area of a regular shape with the same base and altitude, but I do not remember the proportion – less than half but not sure exactly what proportion – 1/3 or ¼? Volume of regular shape = Base * Height = 50 cm^2 * 60 cm = 3,000 cm^3 I will guess it is 1/3 the volume of the regular shape with the same base size, therefore Pyramid Volume = 1/3 * 3000 cm^3 = 1000 cm^3 Confidence Assessment: 2
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Given Solution: `aWe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guessed right on the proportion (1/3 vs ¼) – it has been decades since I used this stuff so not too bad. I will put that one in my notes to review. Self-critique Rating: 3 ********************************************* Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters? ********************************************* Your solution: I do remember the proportion was the same for the cone to a cylinder with like area base, as for the pyramid to the rectangle with like area base, so: Cone Volume = Base*Altitude*1/3= 20m^2 * 9m * 1/3 = 60m^3 Confidence Assessment:3
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Given Solution: `aJust 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating:3 ********************************************* Question: `q008. What is a volume of a sphere whose radius is 4 meters? ********************************************* Your solution: No clue – don’t remember this formula at all. Guess: it is based on a circle so I am sure pi is in there somewhere, and volume, based on a circle so we probably have a r^3 somewhere? Where is the multiple choice question when I need it? ;-) Confidence Assessment: 0
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 4/3 pi r^3 Write this formula down and study as well. Applying the formula is clear. Self-critique Rating: ********************************************* Question: `q009. What is the volume of a planet whose diameter is 14,000 km? ********************************************* Your solution: OK, assuming a nice normal (non-Sci-Fi) sphere type planet Diameter = 14,000 km = 2* r r=7,000km Volume = 4/3*pi*r^3 = 4/3 * pi * (7,000 km)^3 = 4/3 * pi* 343,000,000,000km^3 = = pi*456.333…*10^9 Confidence Assessment:
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions? ********************************************* Your solution: Any regular shape (one with a consistent cross section) having three dimensions will have a volume = area of the base (cross section) * altitude. The shape need not be symmetrical for this to be valid. Any base (triangle, star, irregular blob) as long as it is consistent as a cross section throughout the shape, will work with this formula. Confidence Assessment: 3
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone? ********************************************* Your solution: Both are 1/3 the volume of the figure with the same base (pyramid to rectangle and cone to cylinder). The calculation is based on the fact that the base is the same, but the cross sections are not. Confidence Assessment: 3
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q012. Summary Question 3: What is the formula for the volume of a sphere? ********************************************* Your solution: V = 4/3*pi*r^3 Confidence Assessment: 3
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Given Solution: `aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: ********************************************* Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Added two formulas to study: Volume of Sphere ( V=4/3*pi*r^3) and Volume of pyramid/cone ratio is 1/3 of volume of regular shape that shares the base area. Self-critique Rating: 3 "