This is ""rates"" and assignments 1,2,3. The previous two will follow in a separate form. Assignments 1 & 2 were the same thing when I ran them so I hope that wasn't just a mistake on my part." "Ɇ«fͨú¦m¤³ÀŸ÷ú¥â‰œù assignment #001 001. Rates qa rates 06-12-2007
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18:42:48 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
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RESPONSE --> I am to proceed through this assignment and critique my answers as I have been doing by using the ""Next Question/Answer"" button and the ""Enter Response."" I also can save anything I type in the right panel as notes by clicking the ""Save as notes"" button. confidence assessment: 3
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18:43:06 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
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RESPONSE --> ok confidence assessment: 3
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18:43:35 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
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RESPONSE --> $10/hr confidence assessment: 3
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18:44:13 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> confidence assessment: 3
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18:44:53 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> $5,000 $60,000/12 = $5,000 confidence assessment: 3
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18:45:14 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> confidence assessment:
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18:46:05 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
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RESPONSE --> The business makes an average of $5000 per month. confidence assessment: 3
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18:46:42 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
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RESPONSE --> OK confidence assessment: 3
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18:48:01 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I didn't solve this problem. I must have misclicked somewhere, but it seems like I would have gotten it right. confidence assessment: 3
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18:49:14 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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RESPONSE --> 1200/60 = 20 mpg confidence assessment: 3
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18:51:49 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I completely misinterpreted this calculation to mean miles per gallon. The explanation clarifies why it would be gallons per mile: the question asks for the rate of fuel usage with respect to miles traveled. confidence assessment: 3
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18:53:17 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
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RESPONSE --> The quantity is given, whether it is gallons of gasoline used, miles driven, months, or years, and therefore does not have to be calculated. confidence assessment: 1
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18:53:45 The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK confidence assessment: 3
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18:56:38 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
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RESPONSE --> (162-147)/(50-10) = 15/40 = 3/8 3/8 pound/daily pushup confidence assessment: 2
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18:56:55 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> Yay! confidence assessment: 3
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18:58:32 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
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RESPONSE --> (188-171)/(30-10) = 17/20 .85 pounds per added pound of weight confidence assessment: 2
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18:58:43 The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK confidence assessment: 3
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18:59:52 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
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RESPONSE --> (200-100)m/(22-12)s 100m/10s 10 m/s confidence assessment: 2
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19:00:12 The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK! confidence assessment:
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19:02:09 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
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RESPONSE --> (200m-100m)/9.5 m/s 100m / 9.5 m/s = 10.5 seconds confidence assessment: 2
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19:02:49 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK confidence assessment: 3
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19:04:22 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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RESPONSE --> We do it now because we're finding the average of two rates to get a rate with the same unit of measurement. confidence assessment: 1
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19:04:48 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I understand. confidence assessment: 3 _________________________________________________________________________ Á±‚ô÷¦Òw†µ°M§¬ÒĬƒµwûûª assignment #001 004. Units of volume measure qa areas volumes misc 06-12-2007
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19:08:20 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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RESPONSE --> 10^3 = 1000 confidence assessment: 2
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19:08:40 The 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.
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RESPONSE --> OK self critique assessment: 3
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19:09:37 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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RESPONSE --> 100^3 = 1,000,000 confidence assessment: 3
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19:11:19 It 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.
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RESPONSE --> I went into it thinking the cube was 1 sq cm. If the cubes are 10 cm it would obviously require 10 in each direction (10^3) to make a one cubic meter structure. This requires 1000 smaller cubes. self critique assessment: 2
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19:12:19 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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RESPONSE --> 1000^2 = 1,000,000 confidence assessment: 2
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19:12:33 It 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.
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RESPONSE --> OK! self critique assessment: 3
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19:13:36 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1000 confidence assessment: 3
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19:15:04 A 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.
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RESPONSE --> I just knew a cubic centimeter and milliliter were the same. The insturctor says that a liter, however, is the volume of a 10 cubic cm cube. self critique assessment: 2
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19:22:21 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1000 confidence assessment: 3
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19:24:14 A 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.
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RESPONSE --> One liter is a cube that's 10 cm on each side, and 10*10*10 = 1000 1000 liters = 1 cubic meter self critique assessment: 2
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19:25:09 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> 100*100*100 = 1,000,000 confidence assessment: 3
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19:25:41 There 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.
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RESPONSE --> OK self critique assessment: 3
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19:26:49 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
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RESPONSE --> 1000 kg confidence assessment: 2
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19:27:02 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.
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RESPONSE --> OK self critique assessment: 3
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19:28:38 `q008. What is the mass of a cubic km of water?
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RESPONSE --> 1000*1000*1000 = 1,000,000,000 kg confidence assessment: 2
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19:30:55 A 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.
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RESPONSE --> The mass was 1000 kg per cubic meter, so I should've multiplied my answer by 1000. self critique assessment: 2
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19:39:21 `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?
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RESPONSE --> I have no idea. confidence assessment: 0
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19:43:59 06-12-2007 19:43:59 5 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.
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NOTES -------> 1 cubic km = (10^3 m)^3 = 10^9 m^3 1 m^3 = 1000 liters 1 km^3 = 10^9 m^3 * 10^3 liters/m^3 = 10^12 liters 10^12 liters/(10^10 liters/day) = 10^2 days = 100- days.
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19:47:44 `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?
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RESPONSE --> s.a. = 4*pi*r^2 = 514457600 sq. km. I DON'T KNOW! confidence assessment: 0
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19:49:26 06-12-2007 19:49:26 The 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.
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NOTES -------> V = a * h
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19:49:49 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> I'm not doing well here. confidence assessment: 3
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19:51:07 06-12-2007 19:51:07 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.
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NOTES -------> We visualize centimeter-sized cubes, and a liter is 10 of these on one side. There are 1000 cubic cm in a liter.
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19:52:15 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> .... confidence assessment: 0
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19:53:13 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.
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RESPONSE --> 10 layers of 10 rows of 10 ten-centimeter cubes leading to 1000 10-cm cubes. 10-cm cube represents a liter, so there are 1000 liters in a cubic meter. self critique assessment: 3 - but i meant to save this as notes.
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19:54:47 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> There are 100 centimeters in a meter and this cubed is 1,000,000 confidence assessment: 1
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19:55:09 One 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.
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RESPONSE --> Thank goodness. self critique assessment: 3
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19:57:01 `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?
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RESPONSE --> There are 1000 meters in a kilometer. 1000*1000*1000 = 1,000,000,000 All three dimensions need to be accounted for. confidence assessment: 2
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19:57:16 A 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.
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RESPONSE --> self critique assessment: 3
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19:58:08 `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 saved some things as notes but honestly I'm just confused. I plan to reread the assignment and solutions after submitting it, however. confidence assessment: 3 ________________________________________________________________________ ü¿ÖËò™Â‡éŠž–„Žd¢Á®Û¸cx assignment #002 004. Units of volume measure qa areas volumes misc 06-12-2007
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20:00:21 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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RESPONSE --> 10*10*10 = 1000 cm^3 confidence assessment: 3
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20:00:32 The 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.
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RESPONSE --> Haven't we done this already? self critique assessment: 3
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20:02:00 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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RESPONSE --> 10*10 cm = 1 m 10*10*10 = 1000 cubes confidence assessment: 3
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20:02:07 It 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.
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RESPONSE --> self critique assessment: 3
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20:05:15 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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RESPONSE --> 1000 m * 1000 m = 1,000,000 m^2 squares confidence assessment: 2
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20:05:33 It 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.
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RESPONSE --> self critique assessment: 3
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20:05:46 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1000 confidence assessment: 3
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20:06:03 A 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.
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RESPONSE --> self critique assessment: 3
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20:09:43 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1 m^3 = 1000 liters confidence assessment: 3
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20:10:22 A 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.
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RESPONSE --> self critique assessment: 3
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20:10:50 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> 100 * 100 * 100 = 1,000,000 confidence assessment: 3
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20:10:57 There 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.
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RESPONSE --> self critique assessment: 3
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20:11:41 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
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RESPONSE --> 1000 kg confidence assessment: 3
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20:11:50 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.
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RESPONSE --> self critique assessment: 3
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20:14:21 `q008. What is the mass of a cubic km of water?
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RESPONSE --> 1,000,000,000,0000 kg? confidence assessment: 1
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20:14:33 06-12-2007 20:14:33 A 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.
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NOTES ------->
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20:14:53 `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?
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RESPONSE --> 100 days confidence assessment: 3
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20:15:00 5 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.
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RESPONSE --> self critique assessment: 3
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20:18:01 `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?
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RESPONSE --> sa = 510,000,000 km^2 v = a*h = 1,020,000,000 km^2 confidence assessment: 3
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20:18:15 The 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.
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RESPONSE --> I hope it's ok for us to use our notes from the last round. self critique assessment: 3
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20:19:08 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> A liter is 10 cubic cm high on one side confidence assessment: 2
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20:19:19 06-12-2007 20:19:19 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.
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NOTES ------->
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20:20:11 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> A cubic meter is 10 10-cm cubes high on a side, and therefore 1 cubic meter = 1000 liters confidence assessment: 3
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20:20:18 06-12-2007 20:20:18 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.
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NOTES ------->
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20:21:17 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> 100 * 100 * 100 confidence assessment: 2
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20:21:30 06-12-2007 20:21:30 One 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.
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NOTES ------->
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20:21:58 `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?
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RESPONSE --> Because you have to account for all dimensions of a cubic kilometer. confidence assessment: 3
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20:22:06 06-12-2007 20:22:06 A 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.
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NOTES ------->
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20:22:24 `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 saved more notes this time through than last time. confidence assessment: 3 __________________________________________________________________________ šüñßõŸ†^õ§ˆ¼¶¤ï÷ûOðáýíê¡“øÙÄ assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 06-12-2007
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20:28:06 `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 --> 2lw + 2wh + 2lh = 2(4)(3) + 2(3)(6) + 2(4)(6) 24 + 36 + 48 = 108 sq. m. confidence assessment: 2
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20:28:29 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 --> This problem requires visual aids or I could never do it. self critique assessment: 3
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20:32:58 `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 --> I don't understand the first part of the question. s.a. ( closed cylinder) = (2*pi*r) x h = (2 * pi * 5) * 12 = 376.8 m^2 confidence assessment: 0
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20:33:25 06-12-2007 20:33:25 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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NOTES ------->
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20:35:02 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> 4 pi r^2 = 4*3.14*(3 cm)^2 = 4 * 3.14 * 9 cm^2 =113.04 cm^2 confidence assessment: 2
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20:36:14 06-12-2007 20:36:14 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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NOTES -------> I didn't change diameter to radius before plugging it into the equation
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20:37:51 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> a^2 + b^2 = c^2 sq rt(106) = 10.3 m confidence assessment: 1
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20:38:08 06-12-2007 20:38:08 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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NOTES ------->
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20:39:52 `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 + b^2 = 6^2 b = sq rt(20) = 4.47 m confidence assessment: 3
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20:40:05 06-12-2007 20:40:05 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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NOTES ------->
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20:41:16 `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 --> 700 g/(4 cm * 7 cm * 12 cm) 700 g/336 cm^3 = 2.08 g/cm^3 confidence assessment: 2
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20:41:38 06-12-2007 20:41:38 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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NOTES -------> Answers differ because of rounding.
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20:46:19 `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 --> V = 4/3 pi r^3 V = 4/3 pi 64 V = 85.33 pi m^3 V = 267.94 m^3 mass = 3,000 kg/m^3 * 267.94 m^3 confidence assessment:
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20:46:44 06-12-2007 20:46:44 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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NOTES ------->
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20:51:50 `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 --> How do you find the average density? confidence assessment: 0
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20:52:59 06-12-2007 20:52:59 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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NOTES -------> Average density = total mass/total volume total mass = (density * volume) + (density * volume)
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20:58:13 `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 --> total mass = 80,700 kg total volume = 30 m^3 average density = total mass/total volume 80,700 / 30 = 2690 kg/cubic meter confidence assessment: 3
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20:58:25 06-12-2007 20:58:25 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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NOTES ------->
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21:01:17 `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 --> V = 1700000 m^2 * .015 m = 25500 m^3 m = dv = 25500 m^3 * 860 kg/m^3 m = 21930000 kg confidence assessment: 2
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21:02:29 06-12-2007 21:02:29 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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NOTES -------> Because I did not write down everything I did I cannot see exactly where I went wrong, but the steps I took were the same so I probably made a mistake punching it into the calculator.
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`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 adding the area of the sides (as a rectangle) and the area of the circles on the two ends. confidence assessment: 2
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05:46:59 06-13-2007 05:46:59 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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NOTES -------> Oh. Acylinder = 2 pi r h + 2 pi r^2
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05:48:37 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> A = 4 pi r^2 confidence assessment: 2
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05:48:46 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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05:49:48 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density describes the mass in a certain amount of volume. confidence assessment: 2
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05:50:04 06-13-2007 05:50:04 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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NOTES ------->
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05:50:39 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> v = m/d confidence assessment: 3
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05:50:47 06-13-2007 05:50:47 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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NOTES ------->
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05:51:09 `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 saved all of these as notes which I will go back and organize shortly. confidence assessment: 3
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"
This is ""rates"" and assignments 1,2,3. The previous two will follow in a separate form. Assignments 1 & 2 were the same thing when I ran them so I hope that wasn't just a mistake on my part." "Ɇ«fͨú¦m¤³ÀŸ÷ú¥â‰œù assignment #001 001. Rates qa rates 06-12-2007
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18:42:48 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
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RESPONSE --> I am to proceed through this assignment and critique my answers as I have been doing by using the ""Next Question/Answer"" button and the ""Enter Response."" I also can save anything I type in the right panel as notes by clicking the ""Save as notes"" button. confidence assessment: 3
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18:43:06 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
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RESPONSE --> ok confidence assessment: 3
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18:43:35 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
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RESPONSE --> $10/hr confidence assessment: 3
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18:44:13 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> confidence assessment: 3
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18:44:53 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> $5,000 $60,000/12 = $5,000 confidence assessment: 3
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18:45:14 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> confidence assessment:
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18:46:05 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
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RESPONSE --> The business makes an average of $5000 per month. confidence assessment: 3
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18:46:42 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
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RESPONSE --> OK confidence assessment: 3
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18:48:01 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I didn't solve this problem. I must have misclicked somewhere, but it seems like I would have gotten it right. confidence assessment: 3
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18:49:14 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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RESPONSE --> 1200/60 = 20 mpg confidence assessment: 3
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18:51:49 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I completely misinterpreted this calculation to mean miles per gallon. The explanation clarifies why it would be gallons per mile: the question asks for the rate of fuel usage with respect to miles traveled. confidence assessment: 3
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18:53:17 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
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RESPONSE --> The quantity is given, whether it is gallons of gasoline used, miles driven, months, or years, and therefore does not have to be calculated. confidence assessment: 1
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18:53:45 The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK confidence assessment: 3
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18:56:38 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
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RESPONSE --> (162-147)/(50-10) = 15/40 = 3/8 3/8 pound/daily pushup confidence assessment: 2
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18:56:55 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> Yay! confidence assessment: 3
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18:58:32 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
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RESPONSE --> (188-171)/(30-10) = 17/20 .85 pounds per added pound of weight confidence assessment: 2
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18:58:43 The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK confidence assessment: 3
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18:59:52 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
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RESPONSE --> (200-100)m/(22-12)s 100m/10s 10 m/s confidence assessment: 2
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19:00:12 The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK! confidence assessment:
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19:02:09 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
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RESPONSE --> (200m-100m)/9.5 m/s 100m / 9.5 m/s = 10.5 seconds confidence assessment: 2
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19:02:49 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> OK confidence assessment: 3
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19:04:22 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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RESPONSE --> We do it now because we're finding the average of two rates to get a rate with the same unit of measurement. confidence assessment: 1
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19:04:48 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I understand. confidence assessment: 3 _________________________________________________________________________ Á±‚ô÷¦Òw†µ°M§¬ÒĬƒµwûûª assignment #001 004. Units of volume measure qa areas volumes misc 06-12-2007
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19:08:20 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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RESPONSE --> 10^3 = 1000 confidence assessment: 2
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19:08:40 The 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.
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RESPONSE --> OK self critique assessment: 3
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19:09:37 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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RESPONSE --> 100^3 = 1,000,000 confidence assessment: 3
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19:11:19 It 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.
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RESPONSE --> I went into it thinking the cube was 1 sq cm. If the cubes are 10 cm it would obviously require 10 in each direction (10^3) to make a one cubic meter structure. This requires 1000 smaller cubes. self critique assessment: 2
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19:12:19 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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RESPONSE --> 1000^2 = 1,000,000 confidence assessment: 2
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19:12:33 It 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.
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RESPONSE --> OK! self critique assessment: 3
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19:13:36 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1000 confidence assessment: 3
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19:15:04 A 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.
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RESPONSE --> I just knew a cubic centimeter and milliliter were the same. The insturctor says that a liter, however, is the volume of a 10 cubic cm cube. self critique assessment: 2
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19:22:21 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1000 confidence assessment: 3
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19:24:14 A 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.
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RESPONSE --> One liter is a cube that's 10 cm on each side, and 10*10*10 = 1000 1000 liters = 1 cubic meter self critique assessment: 2
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19:25:09 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> 100*100*100 = 1,000,000 confidence assessment: 3
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19:25:41 There 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.
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RESPONSE --> OK self critique assessment: 3
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19:26:49 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
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RESPONSE --> 1000 kg confidence assessment: 2
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19:27:02 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.
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RESPONSE --> OK self critique assessment: 3
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19:28:38 `q008. What is the mass of a cubic km of water?
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RESPONSE --> 1000*1000*1000 = 1,000,000,000 kg confidence assessment: 2
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19:30:55 A 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.
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RESPONSE --> The mass was 1000 kg per cubic meter, so I should've multiplied my answer by 1000. self critique assessment: 2
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19:39:21 `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?
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RESPONSE --> I have no idea. confidence assessment: 0
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19:43:59 06-12-2007 19:43:59 5 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.
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NOTES -------> 1 cubic km = (10^3 m)^3 = 10^9 m^3 1 m^3 = 1000 liters 1 km^3 = 10^9 m^3 * 10^3 liters/m^3 = 10^12 liters 10^12 liters/(10^10 liters/day) = 10^2 days = 100- days.
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19:47:44 `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?
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RESPONSE --> s.a. = 4*pi*r^2 = 514457600 sq. km. I DON'T KNOW! confidence assessment: 0
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19:49:26 06-12-2007 19:49:26 The 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.
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NOTES -------> V = a * h
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19:49:49 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> I'm not doing well here. confidence assessment: 3
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19:51:07 06-12-2007 19:51:07 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.
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NOTES -------> We visualize centimeter-sized cubes, and a liter is 10 of these on one side. There are 1000 cubic cm in a liter.
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19:52:15 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> .... confidence assessment: 0
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19:53:13 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.
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RESPONSE --> 10 layers of 10 rows of 10 ten-centimeter cubes leading to 1000 10-cm cubes. 10-cm cube represents a liter, so there are 1000 liters in a cubic meter. self critique assessment: 3 - but i meant to save this as notes.
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19:54:47 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> There are 100 centimeters in a meter and this cubed is 1,000,000 confidence assessment: 1
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19:55:09 One 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.
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RESPONSE --> Thank goodness. self critique assessment: 3
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19:57:01 `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?
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RESPONSE --> There are 1000 meters in a kilometer. 1000*1000*1000 = 1,000,000,000 All three dimensions need to be accounted for. confidence assessment: 2
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19:57:16 A 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.
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RESPONSE --> self critique assessment: 3
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19:58:08 `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 saved some things as notes but honestly I'm just confused. I plan to reread the assignment and solutions after submitting it, however. confidence assessment: 3 ________________________________________________________________________ ü¿ÖËò™Â‡éŠž–„Žd¢Á®Û¸cx assignment #002 004. Units of volume measure qa areas volumes misc 06-12-2007
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20:00:21 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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RESPONSE --> 10*10*10 = 1000 cm^3 confidence assessment: 3
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20:00:32 The 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.
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RESPONSE --> Haven't we done this already? self critique assessment: 3
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20:02:00 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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RESPONSE --> 10*10 cm = 1 m 10*10*10 = 1000 cubes confidence assessment: 3
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20:02:07 It 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.
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RESPONSE --> self critique assessment: 3
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20:05:15 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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RESPONSE --> 1000 m * 1000 m = 1,000,000 m^2 squares confidence assessment: 2
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20:05:33 It 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.
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RESPONSE --> self critique assessment: 3
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20:05:46 `q004. How many cubic centimeters are there in a liter?
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RESPONSE --> 1000 confidence assessment: 3
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20:06:03 A 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.
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RESPONSE --> self critique assessment: 3
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20:09:43 `q005. How many liters are there in a cubic meter?
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RESPONSE --> 1 m^3 = 1000 liters confidence assessment: 3
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20:10:22 A 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.
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RESPONSE --> self critique assessment: 3
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20:10:50 `q006. How many cm^3 are there in a cubic meter?
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RESPONSE --> 100 * 100 * 100 = 1,000,000 confidence assessment: 3
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20:10:57 There 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.
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RESPONSE --> self critique assessment: 3
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20:11:41 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
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RESPONSE --> 1000 kg confidence assessment: 3
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20:11:50 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.
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RESPONSE --> self critique assessment: 3
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20:14:21 `q008. What is the mass of a cubic km of water?
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RESPONSE --> 1,000,000,000,0000 kg? confidence assessment: 1
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20:14:33 06-12-2007 20:14:33 A 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.
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NOTES ------->
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20:14:53 `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?
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RESPONSE --> 100 days confidence assessment: 3
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20:15:00 5 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.
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RESPONSE --> self critique assessment: 3
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20:18:01 `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?
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RESPONSE --> sa = 510,000,000 km^2 v = a*h = 1,020,000,000 km^2 confidence assessment: 3
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20:18:15 The 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.
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RESPONSE --> I hope it's ok for us to use our notes from the last round. self critique assessment: 3
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20:19:08 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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RESPONSE --> A liter is 10 cubic cm high on one side confidence assessment: 2
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20:19:19 06-12-2007 20:19:19 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.
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NOTES ------->
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20:20:11 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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RESPONSE --> A cubic meter is 10 10-cm cubes high on a side, and therefore 1 cubic meter = 1000 liters confidence assessment: 3
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20:20:18 06-12-2007 20:20:18 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.
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20:21:17 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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RESPONSE --> 100 * 100 * 100 confidence assessment: 2
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20:21:30 06-12-2007 20:21:30 One 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.
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20:21:58 `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?
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RESPONSE --> Because you have to account for all dimensions of a cubic kilometer. confidence assessment: 3
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20:22:06 06-12-2007 20:22:06 A 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.
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20:22:24 `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 saved more notes this time through than last time. confidence assessment: 3 __________________________________________________________________________ šüñßõŸ†^õ§ˆ¼¶¤ï÷ûOðáýíê¡“øÙÄ assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 06-12-2007
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20:28:06 `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 --> 2lw + 2wh + 2lh = 2(4)(3) + 2(3)(6) + 2(4)(6) 24 + 36 + 48 = 108 sq. m. confidence assessment: 2
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20:28:29 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 --> This problem requires visual aids or I could never do it. self critique assessment: 3
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20:32:58 `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 --> I don't understand the first part of the question. s.a. ( closed cylinder) = (2*pi*r) x h = (2 * pi * 5) * 12 = 376.8 m^2 confidence assessment: 0
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20:33:25 06-12-2007 20:33:25 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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20:35:02 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> 4 pi r^2 = 4*3.14*(3 cm)^2 = 4 * 3.14 * 9 cm^2 =113.04 cm^2 confidence assessment: 2
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20:36:14 06-12-2007 20:36:14 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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NOTES -------> I didn't change diameter to radius before plugging it into the equation
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20:37:51 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> a^2 + b^2 = c^2 sq rt(106) = 10.3 m confidence assessment: 1
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20:38:08 06-12-2007 20:38:08 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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20:39:52 `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 + b^2 = 6^2 b = sq rt(20) = 4.47 m confidence assessment: 3
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20:40:05 06-12-2007 20:40:05 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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20:41:16 `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 --> 700 g/(4 cm * 7 cm * 12 cm) 700 g/336 cm^3 = 2.08 g/cm^3 confidence assessment: 2
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20:41:38 06-12-2007 20:41:38 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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NOTES -------> Answers differ because of rounding.
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20:46:19 `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 --> V = 4/3 pi r^3 V = 4/3 pi 64 V = 85.33 pi m^3 V = 267.94 m^3 mass = 3,000 kg/m^3 * 267.94 m^3 confidence assessment:
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20:46:44 06-12-2007 20:46:44 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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20:51:50 `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 --> How do you find the average density? confidence assessment: 0
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20:52:59 06-12-2007 20:52:59 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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NOTES -------> Average density = total mass/total volume total mass = (density * volume) + (density * volume)
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20:58:13 `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 --> total mass = 80,700 kg total volume = 30 m^3 average density = total mass/total volume 80,700 / 30 = 2690 kg/cubic meter confidence assessment: 3
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20:58:25 06-12-2007 20:58:25 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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21:01:17 `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 --> V = 1700000 m^2 * .015 m = 25500 m^3 m = dv = 25500 m^3 * 860 kg/m^3 m = 21930000 kg confidence assessment: 2
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21:02:29 06-12-2007 21:02:29 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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NOTES -------> Because I did not write down everything I did I cannot see exactly where I went wrong, but the steps I took were the same so I probably made a mistake punching it into the calculator.
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`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 adding the area of the sides (as a rectangle) and the area of the circles on the two ends. confidence assessment: 2
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05:46:59 06-13-2007 05:46:59 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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NOTES -------> Oh. Acylinder = 2 pi r h + 2 pi r^2
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05:48:37 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> A = 4 pi r^2 confidence assessment: 2
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05:48:46 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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05:49:48 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density describes the mass in a certain amount of volume. confidence assessment: 2
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05:50:04 06-13-2007 05:50:04 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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05:50:39 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> v = m/d confidence assessment: 3
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05:50:47 06-13-2007 05:50:47 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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05:51:09 `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 saved all of these as notes which I will go back and organize shortly. confidence assessment: 3
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