#$&* course Phy 201
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Given Solution: `aA rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you need to find the circumference. This cylander’s circumference is 2pi r or 10 pi cm^2. The multiply the circumference by the altitude 10* 12= 120 pi cm^2. To figure out the surface area if it were closed you would take the circumference times the area of the base ( A= pi r^2 or A=pi 5^2=A=25 ) so it would be 120 pi cm^2 by 2+25 pi cm^2=170 pi m^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q003. What is surface area of a sphere of diameter three cm? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 pi r^2 or 4 pi (1.5)^2= 9 pi cm^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2. NOTE TO STUDENT: While your work on most problems has been good, you left this problem blank and didn't self-critique. You should self-critique here. • For example you should acknowledge having made note of the formula for the surface area of the sphere, which I expect you didn't know before. I expect from your previous answers that you are very capable of applying the formula once you have it, and based on this history you probably wouldn't need to self-critique that aspect of the process. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5^2+9^2= 106 m^2, then take the square root to find the hypotenuse- 10.2956 or 10.3 is the hypotenous. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for a hypotenuse is H (c )= sqrt of a^2 + b^2. This can also be c^2=a^2+b^2. Threfore 6^2=4^2+b^2= 36=16+b^2= 20=b^2 or sqrt of 20 or 4.47. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aIf c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My numbers were a little off, but not by too much, I followed the same formula but my rounding was probably a little different. ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we need to find the volume which is 4*7*12 or 336 cm. To find the density you would divide the mass by the volume= 700/336= 2.08 cm^3. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe 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 (for example 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). NOTE TO STUDENT: (in this note the instructor attempts to clarify the idea of 'demonstrating what you do and do not understand about the statement of the problem' and 'giving a phrase-by-phrase analysis of the given solution') You did not respond to the question and did not self-critique. You would be expected to address the question, stating what you do and do not understand. • For example you should understand what a rectangular solid with dimensions 4 cm by 7 cm by 12 cm is, and how to find its volume and surface area. You might not know what to do with this information (for example you might well not understand that it's the volume and not the surface area that's related to density), but from previous work you should understand this much, and should at least mention something along the lines of 'well, I do know that I can find the volume and/or surface area of that solid' in a partial solution. • The word 'density' is clearly very important. Even if you don't know what density is, you could note from the statement of the problem that its units here are said to be 'grams per cubic centimeter'. Having noted these things, you will be much better prepared to understand the information in the given solution. Then you need to address the information in the given solution. A 'phrase-by-phrase' analysis is generally very beneficial: • I expect you understand the first statement from previous knowledge (you should have this understanding from prerequisite courses, and if not you encountered it in the preceding 'volumes' exercise): 'The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.' It would of course be appropriate to ask a question here if necessary. • It is likely that, as is the case with many students, the concept of density is not that familiar to you. However if this wasn't addressed specifically in prerequisite courses, those courses would be expected to prepare you to understand this concept. The statement 'Its density in grams per cm^3 is the number of grams in each cm^3.' serves as a definition of density. In your self-critique you should have addressed what what this phrase means to you, and what you do or do not understand about it • The next phrase is 'We find this quantity by dividing the number of grams by the number of cm^3.' You would be expected to understand that this phrase is related to the preceding, and as best you can to address the connection. At this point many students would need to ask a question, and it would be perfectly appropriate to do so (or to have done so regarding previous statements). • The subsequent phrase 'density = 700 grams / (336 cm^3) = 2.06 grams / cm^3' is an illustration of the ideas and definitions in the preceding statements. A reasonable self-critique would demonstrate your attempt to understand this statement and its connection to the preceding. Once again questions would also be appropriate and welcome. • The above addresses sufficient information to solve the problem. If you get to this point, you're probably doing OK and you wouldn't necessarily be expected to address the rest of the given solution, which expands on the finer details of the problem and provides additional information. The basic prerequisite courses should have prepared you to understand the information, but students entering Liberal Arts Mathematics, College Algebra and even Precalculus or Applied Calculus (or Physics 121-122) courses probably don't need to address anything beyond the basic solution at this point. Though Precalculus and Applied Calculus students could benefit from doing so, and if time permits would certainly be encouraged to do so, time is also a factor and it would be understandable if these students chose to move on. • Students entering the Mth 173-4 sequence or the Phy 201-202 or 231-232 sequence would be expected to either completely understand all the details of the given solution, or address them in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We first need to find the V which is 4/3pi r^3 or 85.3. Then Mass = D*V so 3,000*85.3=255,900 m^3 . This would be the mass. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understood how to get to the answer, but this answer was rounded up completely whereas mine was not. ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We first need to find the mass of these objects or density * volume. 6*4= 24 grams per cm^3. The other mass is 10*2 or 20 cm^2. The average density would be the total mass divided by volume. This would equal 24+20=44 divided by 16(total volume)=2.75 cm^3 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the average density we first have to find the total mass of the sand and the cannonball. The total mass of the sand is 27*2100=56,700. The total mass of the cannon ball is 3*8000=24,000. The avg density is total mass/total volume. The total mass is 807,000/ (27+3)30=2,690 m^3. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aWe find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not round my answer up but we arrived at a similar answer in the same way. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the cubic meters I multiplied the 1700000 by the depth .015 which gives us 25,500m^3. Density is mass*volume so we would do 860= mass*volume or Mass= 860*volume. 860*25,500=21,930,000m ^3 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 25,500 m^3 = 21 930 000 kg. This result should be rounded according to the number of significant figures in the given information. STUDENT QUESTION I didn’t round to the most significant figure. ???? How important is this? INSTRUCTOR RESPONSE It will be important. This document is preliminary; the issue of significant figures will be addressed more specifically as we move into the course. Right now I just want you to be aware of the general idea. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the surface area of a cylander you multiply the circumference times the altitude. This determines the sides surface area so next you have to find the surface area for the top and bottom by adding the previous to 2 pi r^2. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aThe curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The surface area is A= pi r^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The term density is the mass divided by the volume. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density' &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer was correct however I could have gone into greater detail. ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since density = mass/volume you can determine volume by dividing mass by density. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q015. Part 1 Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. I have added these notes for reference. "
#$&* course Phy 201 004. Units of volume measure
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Given Solution: `aThe volume of the container is 10 cm * 10 cm * 10 cm = 1000 cm^3. So it would take 1000 cubic centimeters of fluid to fill the container. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A meter has 1000cm to it so it would require 1000 cubes to build a solid cube. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aIt takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high. It should therefore be clear that the large cube could be built using 10 layers, each consisting of 10 rows of 10 small cubes. This would require 10 * 10 * 10 = 1000 of the smaller cubes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would take 1,000,000 square tiles. This is because there are 1,000 meters to a kilometer so in order to make it squared tiles you would do 1,000*1000. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aIt takes 1000 meters to make a kilometer (km). To cover a square 1 km on a side would take 1000 rows each with 1000 such tiles to cover 1 square km. It therefore would take 1000 * 1000 = 1,000,000 squares each 1 m on a side to cover a square one km on a side. We can also calculate this formally. Since 1 km = 1000 meters, a square km is (1 km)^2 = (1000 m)^2 = 1,000,000 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I need to remember to add the proper units after my answer (m^2). ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q004. How many cubic centimeters are there in a liter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There would be 1,000 cm in a liter. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA liter is the volume of a cube 10 cm on a side. Such a cube has volume 10 cm * 10 cm * 10 cm = 1000 cm^3. There are thus 1000 cubic centimeters in a liter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I could have provided more explanation for this but I ended up with the same answer. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q005. How many liters are there in a cubic meter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since it would take ten cm to fill up a liter it would take about 1,000 liters to fill up a cubic meter. (10*10*10) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aA liter is the volume of a cube 10 cm on a side. It would take 10 layers each of 10 rows each of 10 such cubes to fill a cube 1 meter on a side. There are thus 10 * 10 * 10 = 1000 liters in a cubic meter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was not entirely sure about the answer to this problem, however I have a better understanding after reading this explanation. ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q006. How many cm^3 are there in a cubic meter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 1,000,000 because there are 100 cm in a meter and if you cube this it would be 1,000,000. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThere are 1000 cm^3 in a liter and 1000 liters in a m^3, so there are 1000 * 1000 = 1,000,000 cm^3 in a m^3. It's important to understand the 'chain' of units in the previous problem, from cm^3 to liters to m^3. However another way to get the desired result is also important: There are 100 cm in a meter, so 1 m^3 = (1 m)^3 = (100 cm)^3 = 1,000,000 cm^3. STUDENT COMMENT It took me a while to decipher this one out, but I finally connected the liters to cm^3 and m^3. I should have calculated it by just converting units, it would have been easier. INSTRUCTOR RESPONSE The point isn't just conversion. There are two points to understanding the picture. One is economy of memory: it's easier to remember the picture than the conversion factors, which can easily be confused. The other is conceptual/visual: the picture gives you a deeper understanding of the units. In the long run it's easier to remember that a liter is a 10-cm cube, and a cubic meter is a 100-cm cube. Once you get this image in your mind, it's obvious how 10 layers of 10 rows of 10 one-cm cubes forms a liter, and 10 layers of 10 rows of 10 one-liter cubes forms a cubic meter. Once you understand this, rather than having a meaningless conversion number you have a picture that not only gives you the conversion, but can be used to visualize the meanings of the units and how they are applied to a variety of problems and situations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 1,000 liters in a cubic meter so in turn the mass would be 1000 kg. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Since there are 1000 liters in a cubic meter, the mass of a cubic meter of water will be 1000 kg. This is a little over a ton. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q008. What is the mass of a cubic km of water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a cubic meter has a mass of 1,000 kg then a cubic km would be 1000^3 so it would be 1,000,000,000 cubic meters times 1,000 kg or 1,000000000000kg. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aA cubic meter of water has a mass of 1000 kg. A cubic km is (1000 m)^3 = 1,000,000,000 m^3, so a cubic km will have a mass of 1,000,000,000 m^3 * 1000 kg / m^3 = 1,000,000,000,000 kg. In scientific notation we would say that 1 m^3 has a mass of 10^3 kg, a cubic km is (10^3 m)^3 = 10^9 m^3, so a cubic km has mass (10^9 m^3) * 1000 kg / m^3 = 10^12 kg. STUDENT QUESTION I don’t understand why you multiplied the 1,000,000,000 m^3 by 1000 km/m^3. I also don’t understand where the (1000m)^3 came from. I thought I had this problem but it stumped me. It is probably something really simple that I am missing. ??? INSTRUCTOR RESPONSE A km is 1000 meters, but a cubic km is a cube 1000 meters on a side. It would take 1000 m^3 just to make a single row of 1-m cubes 1000 meters long, and you would hardly have begun constructing a cubic kilometer. You would need 1000 such rows just to cover a 1-km square 1 meter deep, and 1000 equal layers to build a cube 1 km high. Each layer would require 1000 * 1000 cubic meters, and 1000 layers would require 1000 times this many 1-meter cubes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t sure I had this problem right but after seeing the answer I see that I did it correctly and did not over think it. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q009. If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would divid 1,000,000,000,000 by the amount of liters people are drinking which would be 10,000,000,000 which would equal 100. So it would take them 100 days. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a5 billion people drinking 2 liters per day would consume 10 billion, or 10,000,000,000, or 10^10 liters per day. A cubic km is (10^3 m)^3 = 10^9 m^3 and each m^3 is 1000 liters, so a cubic km is 10^9 m^3 * 10^3 liters / m^3 = 10^12 liters, or 1,000,000,000,000 liters. At 10^10 liters per day the time required to consume a cubic km would be time to consume 1 km^3 = 10^12 liters / (10^10 liters / day) = 10^2 days, or 100 days. This calculation could also be written out: 1,000,000,000,000 liters / (10,000,000,000 liters / day) = 100 days. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q010. The radius of the Earth is approximately 6400 kilometers. What is the surface area of the Earth? If the surface of the Earth was covered to a depth of 2 km with water that what would be the approximate volume of all this water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A=4pi r^2 or 4 pi *6400 km^2 or 510,000,000 pi km^3. Surface = A* H= 510,000,000 pi km^3*2 = 1,020,000,000 km ^3 is the volume. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `aThe surface area would be A = 4 pi r^2 = 4 pi ( 6400 km)^2 = 510,000,000 km^2. A flat area of 510,000,000 km^2 covered to a depth of 2 km would indicate a volume of V = A * h = 510,000,000 km^2 * 2 km = 1,020,000,000 km^3. However the Earth's surface is curved, not flat. The outside of the 2 km covering of water would have a radius 2 km greater than that of the Earth, and therefore a greater surface area. However a difference of 2 km in 6400 km will change the area by only a fraction of one percent, so the rounded result 1,020,000,000,000 km^3 would still be accurate. STUDENT COMMENT I thought that in general pi was always supposed to be expressed as pi when not asked for an approximate value so in the first part of the problem I didn’t calculate pi. For the second part of the question I assumed approximate meant calculate pi into the equation which would still be a less precise answer so I did not round any further. ???Should I have estimated more than I did??? INSTRUCTOR RESPONSE The given information says 'approximately 6400 km'. Your result, 163,840,000pi km^2, is perfectly fine. However most people aren't going to recognize 163,840,000 as 4 times the square of 6400 (unlike a result like 36 pi (easily enough seen as either 6^2 * pi, or 4 * 3^2 * pi)). Since the given information is accurate to only a couple of significant figures, there's no real advantage in the multiple-of-pi expression. In the given solution the results are generally expressed to 2 significant figures, consistent with the 2 significant figures in the given 6400 km radius. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Although I got this correct, it gave me a better understanding after seeing this explanation. ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We visualize by thinking of layers of 10 cubes so this would be about 1,000 cubic centimeters. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: Since a liter is a cube 10 cm on a side, we visualize 10 layers each of 10 rows each of 10 one-centimeter cubes, for a total of 1000 1-cm cubes. There are 1000 cubic cm in a liter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is similar to the last question as the last since there are 10 cubes to cover a meter. This would equal about 1,000 again. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Since a liter is a cube 10 cm on a side, we need 10 such cubes to span 1 meter. So we visualize 10 layers each of 10 rows each of 10 ten-centimeter cubes, for a total of 1000 10-cm cubes. Again each 10-cm cube is a liter, so there are 1000 liters in a cubic meter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In this question it takes 100 cm to fill a meter and if this is cubic then it would be 100^3= 100,000. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aOne way is to know that there are 1000 liters in a cubic meters, and 1000 cubic centimeters in a cubic meter, giving us 1000 * 1000 = 1,000,000 cubic centimeters in a cubic meter. Another is to know that it takes 100 cm to make a meter, so that a cubic meter is (100 cm)^3 = 1,000,000 cm^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q014. Summary Question 4: There are 1000 meters in a kilometer. So why aren't there 1000 cubic meters in a cubic kilometer? Or are there? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A cubic meter would be 1000^3 which would equal 1000,000,000. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aA cubic kilometer is a cube 1000 meters on a side, which would require 1000 layers each of 1000 rows each of 1000 one-meter cubes to fill. So there are 1000 * 1000 * 1000 = 1,000,000,000 cubic meters in a cubic kilometer. Alternatively, (1 km)^3 = (10^3 m)^3 = 10^9 m^3, not 1000 m^3. STUDENT ANSWER to question: Because a cubic kilometer is cubed. A regular kilometer is not going to contain as much as a cubic kilometer. INSTRUCTOR RESPONSE Kilometers and cubic kilometers don't measure the same sort of thing, so they can't be compared at all. Kilometers measure distance, how far it is between two points. Cubic kilometers measure volume, how much space there is inside of something (there is space, though not necessarily empty space, inside of any container or any 3-dimensional region, whether it's full of other stuff or not. If it's full of other stuff then we wouldn't say that it's 'empty space' or 'available space', but the amount of space inside is the same either way). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment. I have added these problems to my notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK "