course Phy 201 םރͲYcassignment #003
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12:29:47 `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?
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RESPONSE --> I think that the second line which contains the points (7,17) to the point (10,29) is steeper than the other line. When I drew the graph it seemed to me that the second was more straight up than the other line was. confidence assessment: 1
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12:31:28 The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction. Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3. Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.
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RESPONSE --> I got the answer correct but I didn't work it out like it states in the answer. I didn't think about figureing out the slope with the points, I just looked at my graph that I made. self critique assessment: 2
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12:44:03 `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.
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RESPONSE --> These values make the expression zero because when you graph the equation the points 2 and -2.5 where the line intersects with the x axis which means the value of both of those points are zero. confidence assessment: 1
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12:45:37 If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero. If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero. The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero. Note that (x-2)(2x+5) can be expanded using the Distributive Law to get x(2x+5) - 2(2x+5). Then again using the distributive law we get 2x^2 + 5x - 4x - 10 which simplifies to 2x^2 + x - 10. However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.
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RESPONSE --> I guess I need to review the math that I have already taken because I seem to have forgotten it all. self critique assessment: 1
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12:59:36 `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?
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RESPONSE --> The values of x for this expression would be 2 and -4. I found this answer by setting each equation to zero and finding the answer to the equation that would make that equation zero. If one part of the equation was zero then it would make the entire equation zero. confidence assessment: 0
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13:05:10 In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0. 3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero. We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**
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RESPONSE --> I did better than I thought since I got 2 and -4 correct. I looked back at my work I saw that the part of the equation x^2-4, I worked it wrong self critique assessment: 2
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13:16:50 `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> The poinst (3,5) and (7,9) forms the trapazoid with the greatest area because the points make that trapezoid bigger than the other one which more spread out. confidence assessment: 0
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13:22:36 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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RESPONSE --> I thought that with the height of the first trapezoid that it would have more area than the second trapezoid. self critique assessment: 2
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13:38:18 * `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph: As we move from left to right the graph increases as its slope increases. As we move from left to right the graph decreases as its slope increases. As we move from left to right the graph increases as its slope decreases. As we move from left to right the graph decreases as its slope decreases.
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RESPONSE --> For the first graph y = x^2 as I move from left to right the graph decreases while the slope increases. The second graph y = 1/x as I move from left to right the graph decreases and its slope decreases. The third graph y = 'sqrt(x) as I move from left to right the graph increases and the slope increases. self critique assessment: 0
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13:40:40 For x = 1, 2, 3, 4: The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate. The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero. Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases. We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate. For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.
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RESPONSE --> I graphed the equation and thats how I got my incorrect answers. self critique assessment: 1
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13:55:01 `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?
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RESPONSE --> The way I solved this equation was I multiplyed 20 by .10 which resulted in 2. Then 2 frogs times 3 months equals 6. Which means at the end of the 3 months there will be 26 frogs. Then if someone wanted to calculate the number of frogs after 300 months you should multiply 300 by 2 which is the number of increased frogs per month. Once you get that answer which is 600. The you would add 20 to 600 to find the population of frogs in the pond after 300 months. confidence assessment: 0
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13:57:22 At the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs. The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic). A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.
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RESPONSE --> I just took the fact that there was 2 frogs per month I didn't think to add to each one. I multiplyed 2 times 3. I see what I did wrong with working out the equation. self critique assessment: 2
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14:04:37 `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?
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RESPONSE --> When I calculated the problem I got 1/1 = 1 , 1/.1 = 10 , 1/.01 = 100 , 1/.001 = 1000. We say that the values of x are approaching zero because the number that we divide 1 by gets smaller and smaller. We might use the numbers .0001 and .00001to continue approaching the number zero. The values of 1/x gets larger as the denominator gets smaller. The graph of y = 1/x when the x values are between 0 and 1 do not touch the x axis. confidence assessment: 0
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14:05:38 If x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1. So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc.. Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere. The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become. The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis. This is what it means to say that the y axis is a vertical asymptote for the graph .
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RESPONSE --> does increasing without bound means that the line never stops?
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14:11:13 * `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?
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RESPONSE --> I solved this equation by solving for v which was 3(5)+9 = 24. Then I inserted the value of v into the formula for the equation E = 800v^2. When I solved E = 800(24^2) E = 800(576) which equals E = 460800. self critique assessment: 0
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14:11:38 For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.
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RESPONSE --> I got one right!! self critique assessment: 2
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14:14:39 * `q009. Continuing the preceding problem, can you give an expression for E in terms of t?
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RESPONSE --> No you can't give the expression for E in terms of t because you use velocity to figure up the the energy. Time is used to find the velocity not energy. self critique assessment: 0
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14:16:17 Since v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here. For further reference, though, note that this expression could also be expanded by applying the Distributive Law:. Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).
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RESPONSE --> I understand what I did wrong I should have inserted the equation for velocity into the the equation needed to solve that question. self critique assessment: 2
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^ߎ̱| assignment #001 001. Rates qa rates 06-02-2007 ˳Ĥzz`V䥩Vߞ assignment #001 001. Rates qa rates 06-02-2007
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14:31:19 `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 --> When the program starts you click on the next question/Answer to get the question to be answered. In the right hand side of the screen you enter in your answer to the question. Then you click on the Enter Response button to submit your answer. Next you click the Next Question/Answer button to get the correct answer to the question. Then in the right hand side you enter your self-critique of your answered compared to the correct answer. You can save your answers by pushing the Save as Notes button to keep your answers for future reference. After you have entered your critique you push enter response and then you click on the Next/Question Answer to proceed to the next question. confidence assessment: 3
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14:31:44 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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14:38:29 `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 --> If someone was made $50 in 5 hr, the the person is getting $10 an hr. confidence assessment: 3
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14:47:30 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 --> ok confidence assessment: 3
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14:48:53 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> You take $60,000 and divide it by 12 for the 12 months in a year and the answer you get is $5,000 per month confidence assessment: 3
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14:49:26 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 --> ok confidence assessment: 3
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14:51:16 `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 --> If $60,000 is made per year by the small business then it would be more appropriate to say that the business makes $5000 per month. confidence assessment: 2
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14:54:32 06-02-2007 14:54:32 Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. 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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NOTES -------> When I solved the question I was thinking the exact amount was $60,000 so that is why I chose the comment that the business would make $5,000 a month. I realize that it would be impossible to say that the amount every month would have the same amount of money.
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15:05:14 `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 --> I divided 300 by 6 to get the answer 50 miles covered per hour. We say the average rate instead of plain rate because the amount of miles in 6 hrs. can change depending on the conditions so it is more of an estimate of the mileage covered. confidence assessment: 1
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15:07:36 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'm not sure how velocities fit into this problem. I got the answer correct but I didn't even think about velocities. confidence assessment: 2
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15:15:40 `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 --> I divided 1200 by 60 and the answer I got for the question is 20 miles per gallon of gasoline. confidence assessment: 2
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15:23:03 06-02-2007 15:23:03 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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NOTES -------> My first mistake was to not pay attention to what the question was really asking. I divided 1200 by 60 when I should have divided 60 by 1200. I'm not sure I do totally understand the phrase ""with respect to"". 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.
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15:28:53 `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 --> We are not adding anything when we come up with answers to average rate problems because rates represent the change between two variables. confidence assessment: 0
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15:32:21 06-02-2007 15:32:21 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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NOTES -------> I'm not sure my answer would be considered correct because I didn't word it like the answer but my thinking was along the same lines. I didn't think of the fact that when we are calucating average rate we have already added up all the other information needed to answer the question.
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15:50:17 `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 --> I have no idea what to do with this question. confidence assessment: 0
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15:55:06 06-02-2007 15:55:06 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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NOTES -------> Did you get 15 by figuring out how much more weight the second group was lifting than the first group. I see where the 40 comes from its because 50 - 10 right?
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16:01:12 `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 --> I took 188 - 171 = 17 and then I took 30 -10 = 20 then I divided 20 by 17 and got the answer 1.18 confidence assessment: 0
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16:03:29 06-02-2007 16:03:29 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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NOTES -------> I couldn't decide if it was right to divide 17 by 20 or 20 by 17. I have to divide the lifting pounds (17) by the added pounds (20) to get the answer of .85
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16:09:04 `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 --> I solved this equation by subtracting meter mark 200 - 100 = 100 then i subtracted the seconds 22 - 12 = 10. I divided 100 by 10 to get the answer 10. confidence assessment: 0
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16:10:49 06-02-2007 16:10:49 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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NOTES -------> The average rate of the runner is an average rate because at differnt positions in his stride the runner would clearly be traveling at slightly different speeds.
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16:15:24 `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 --> 200 - 100 = 100 and 10 - 9 = 1 So I took the 100 meter mark and divided it by 1 meter/second and go the answer 10. confidence assessment: 0
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16:20:36 06-02-2007 16:20:36 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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NOTES -------> To get the simpliest estimates we could make that the average speed is the average of 10 m/s and 9 m/s. 10 + 9 = 19 and then you divide by 2 to get the answer of 9.5 m/s. Next you would take the 100 meters and divide them by 9.5 m/s to get the answer 10.5 seconds approximatly.
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16:22:16 `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 have to divide by 2 because the average rate of speed was not specified in the problem. So you have to divide by 2 to get an average rate of speed. confidence assessment: 1
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16:22:38 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 --> ok confidence assessment: 2
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Lw°~Ɋ assignment #001 001. Areas qa areas volumes misc 06-02-2007
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16:28:11 `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters.
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RESPONSE --> To get area you multiply 4 by 3 and the awnser that I got was 12. confidence assessment: 2
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16:30:20 06-02-2007 16:30:20 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.
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NOTES -------> Area of a rectangle is found by multiplying the lenght times the width. Unit m stands for meters and m * m = m^2
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16:33:14 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> I used the equation A = 1/2bh to solve this equation. 1/2(4.0)(3.0) = 1/2(12) = 6 The area of the right triangle is 6.
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16:34:17 06-02-2007 16:34:17 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters. The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.
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NOTES -------> Formula for the area of a right triangle with base b and altitude h is A = 1/2 *b*h
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16:35:40 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> To answer this question I multiplyed 5.0 by 2.0 to get the answer 10. The area of this parallelogram is 10. confidence assessment: 1
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16:36:15 06-02-2007 16:36:15 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.
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NOTES -------> Area for a parallelogram is A = b * h
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16:37:36 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> A =1/2*5.0*2.0 and the answer that I got for this question is 5. confidence assessment: 1
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16:37:50 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> ok self critique assessment: 3
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16:41:50 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> To find the area of a trapazoid I multiplyed 1/2 by 5.0 the height of the trapazoid to get the answer 2.5. then I multiply 2.5 by 4.0 + 5.0 to get the answer 22.5. confidence assessment: 1
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16:44:10 06-02-2007 16:44:10 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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NOTES -------> Any trapezoid can be reconstructed to form a rectangle whose width is equal to the average of the two altitudes of the trapezoid so the formula is A = base * average altitude.
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16:47:07 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> to solve this question I will multiply 4 by 1/2 to get the answer 2. I then add up the altitudes 3.0 and 8.0 to get the answer 11. Then I take 2 and multiply it by 11 and the final answer is 22. confidence assessment: 1
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16:47:59 06-02-2007 16:47:59 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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NOTES -------> The area of a trapezoid is equal to the product of the width and the average altitude.
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16:51:01 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> For the area of a circle I will square 3.00 to get 9. I then take 9 and multiply it times pie and I get the answer 28.27 confidence assessment: 1
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16:55:59 06-02-2007 16:55:59 The area of a circle is A = pi * r^2, where r is the radius. Thus A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.
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NOTES -------> When using the area of a circle formula A = pi r^2 gives area in square units and the formula C = 2 pi r gives a result which is in units of radius rather than square units.
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16:57:57 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> The circumference of a circle whose radius is 3 is solved by multiplying 2 * pi * 3 and I got the answer 18.85 confidence assessment: 1
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16:58:49 06-02-2007 16:58:49 The circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.
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NOTES -------> Circumference of a circle C = 2 pi r
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17:02:51 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> To solve this equation in divided the diameter of 12 by 2 to get the radius of the circle which is 6. Then I used the formula A = pi r^2 A = pi * 36= 113.10 confidence assessment: 1
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17:03:48 06-02-2007 17:03:48 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.
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NOTES -------> When finding the area of a circle and only the diameter is given you should divide the diameter in half to get the radius
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17:11:27 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> To find the area of a circle with the circumferencce I think you multiply pi * 14 and I got the answer 43.1 confidence assessment: 1
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17:13:41 06-02-2007 17:13:41 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r. We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.
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NOTES -------> Circumference and radius are related by C = 2 pi r. solving for r we obtain r = C / (2 pi) when you find r you find the area by A = pi r^2
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17:19:12 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> I don't know how to set this question up to be able to solve it. confidence assessment: 0
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17:26:37 06-02-2007 17:26:37 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.
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NOTES -------> Finding the radius of a circle with the area. A = pi r^2 we solve for r. Divide both sides by pi to get A/ pi = r ^2. then reverse the sides and take the square root of both sides obtaining r = sqrt( A/ pi)
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17:28:55 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> When I visualize a rectangles area I think of the height times the width. confidence assessment: 1
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17:30:33 06-02-2007 17:30:33 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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NOTES -------> Visualize the rectangle being covered by rows of 1- unit squares. Multiply the number of squares in a row by the number of rows getting the area A = L * W
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17:34:51 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> We can visualize the area of a triangle by using the formula A = 1/2*b*h confidence assessment: 1
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17:40:29 06-02-2007 17:40:29 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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NOTES -------> To visualize right triangles we think of two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle
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17:46:04 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> We calculate the area of a parallelogram by using the formula A = b*h. Where b is the base and h is the height. confidence assessment: 2
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17:46:15 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> self critique assessment:
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17:46:20 06-02-2007 17:46:20 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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NOTES ------->
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17:47:23 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> We calculate the area of a trapazoid with the formula A = 1/2 * h * (a+b) confidence assessment: 2
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17:48:49 06-02-2007 17:48:49 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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NOTES -------> When calculating a trapezoid we think of its two parallel sides are verical and we multiply the average altitude by the width.
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17:49:44 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> When calculating the area of a circle we multiply pie by the radius squared A = pi * r^2 confidence assessment: 1
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17:50:22 06-02-2007 17:50:22 We use the formula A = pi r^2, where r is the radius of the circle.
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NOTES -------> Circle area A = pi r^2
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17:53:10 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> We calculate the circumference of a circle by multiplying C = 2*pi*r. We can easily avoid getting this confused with the area of a circle but if we remember that we square the radius when finding the area and we don't multiply by 2. confidence assessment: 1
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17:53:21 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> self critique assessment:
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17:57:16 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have organized the principles that were discussed in this assignment by the different shapes that was discussed. Each shape has its own formula to use for the answer to the question. Also I have made a mental note that most of the shapes we discussed have to do with base and height and with radius when working with circles. confidence assessment: 1
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20:25:53 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> To work this problem I have subtracted 30 from 20 to get 10. Take 10 and divide it by 2 gives the answer that it takes 5 seconds for the speedometer to move from 20 mph to 30 mph. To figure out what the speedometer will read 7 seconds later is found by multiplying 7 times 2 which gives you 14. You then add 14 to 10 to give the final answer of 24 mph. confidence assessment: 1
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20:26:15 It will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.
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RESPONSE --> ok self critique assessment: 3
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20:31:40 06-02-2007 20:31:40 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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NOTES ------->
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20:32:37 06-02-2007 20:32:37 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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NOTES ------->
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20:33:03 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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RESPONSE --> I am totally lost on this question,. self critique assessment: 1
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20:39:51 `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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RESPONSE --> I have no clue confidence assessment: 0
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20:44:48 The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. }{The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.
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RESPONSE --> ok self critique assessment: 0
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20:51:58 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> To solve this question I took 3000 and divided it by 1500 which resulted in the answer 2. Then I took 5000 and divided it by 2000 and got the answer 2.5. So the team that had the net force of 5000 Newtons is the team that will win. If someone pulled with force of 500 Newtons then the other team would win. self critique assessment: 0
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20:56:42 The first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team
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RESPONSE --> I got the first part of the question right because I answered that the second team would win but I got the second part of the question wrong and even after the answer stated I'm not sure that I understand the reasoning. I thought that if the resistance was 500 instead of 5000 that the other team would win for sure. self critique assessment: 1
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21:21:21 `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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RESPONSE --> To solve this problem I divided 250/10 = 25 and 200/20 = 10. The player that will be moving backwards immediatly after the collision will be the 200-lb player. confidence assessment: 0
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21:23:32 06-02-2007 21:23:32 Greater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.
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NOTES -------> If we multiply speed by mass we get the determining quantity, which is called momentum.
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21:34:46 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> For this question I solved it by multilplying 200 * 12 = 24600 and 150 * 10 = 1500. So I think that the 200-lb climber should be able to make it further up the mountain. confidence assessment: 1
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21:40:35 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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RESPONSE --> I think part of my problem with these problems is that I am using the wrong operations. In my answer I multiplied 200 * 12 instead of 12 / 200. I see that the second climber has more energy per pound of body weight considering the caculations but I was thinking that if someone was bigger and ate more then they would have more energy. self critique assessment: 1
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ƿߧǃ assignment #006 006. Physics qa initial problems 06-03-2007
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07:51:09 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> For the first question I got the answer 20. I got this by subtracting 30 - 20 = 10 then I took 10 and multiplied it by 2 to get the answer 20. For the second question I multiplied 7 * 2 = 14 I took the 14 and added it to 10 to get the final answer 24. confidence assessment: 1
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07:51:35 It will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.
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RESPONSE --> ok self critique assessment: 3
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08:04:26 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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RESPONSE --> It will require less than 10 seconds to to reach the lamppost because the faster you are going the less time it will take to get to your destination. Since the initial speed was 10 mph. greater than before, I think that the speed at the lamppost would be 10 mph greater than before because different conditions can effect the speed of the car and it is unlikely that the speed of the car would stay the same from lampost to lamppost confidence assessment: 1
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08:08:42 06-03-2007 08:08:42 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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NOTES -------> If velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at, and it would therefore take less time. equal distances, implies less time on the second run. If speed is changing at the same rate as before and it has less time to change it will therefore change by less
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08:26:28 `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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RESPONSE --> The automobile that speeds up from 40 mph to 90 mph in 20 seconds is speeding up at a greater rate because 20 seconds minus 5 seconds equals 15 seconds. 30 mph minus 20 gives me 10 mph difference. 90 mph minus 40 mph is 50 mph differnce. So the second car inceases its speed at a greater rate than the first car. confidence assessment: 1
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08:28:24 06-03-2007 08:28:24 The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. }{The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.
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NOTES -------> When decideing which car will be going at a greater rate of speed you need to divide the mph by the seconds to get the rate of change.
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08:35:30 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> The second team will win the competition of who can most quickly accelerate their initially stationary automobile to 5 mph. First team calculations is 3000/1500 = 2 , 2*5 = 10 the second team 5000/2000 = 2.5 , 2.5*5 = 12.5 No the other team would still not win the competition because 500 Newtons is not enough force to change the results of the competition. self critique assessment: 2
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08:38:01 06-03-2007 08:38:01 The first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team
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NOTES -------> I should divide the Newtons by the amount of kg but you should not multiply the answer from the Newtons and kg's at the end of my calculation.
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08:43:17 `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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RESPONSE --> I think the 250-lb football player will be the player that will be moving backward immediately after the collision because the 200-lb player will have more monentum behind him because he has 10 feet per second more to get going. confidence assessment: 1
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08:44:54 06-03-2007 08:44:54 Greater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.
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NOTES -------> to get momentum we need to multiply speed by the mass to get the answer. We don't use pounds as units.
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08:49:57 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> I think the climber who is 200lb will be the able to climb further up the mountain because 200*12 = 2400. The 150lb climber would be 150*10 = 1500. So the 200 lb climber will do better than the other climber. confidence assessment: 1
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08:53:11 06-03-2007 08:53:11 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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NOTES -------> When finding the number of ounces of Cheerios per pound of body weight. You divide the ounces by the pounds of body weight. The Cheerios supply energy to lift the pounds of climber so the climber with the lowest weight will have climb further than the other climber with more body weight.
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09:07:40 `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop. Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long? Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great? Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?
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RESPONSE --> The automobile that is going faster will take more time to come to a stop because the car is going faster than the other car even though they coast at the same time the greater speed gives the car more momentum which makes the car go farther. The coasting velocity will be more than twice as great as the other. The automobile going faster will have the greater average coasting velocity because the car has more momentum behind it. The average coasting velocity will be twice as great as the other. The distance traveled by the faster automobile will be twice that of the slower vehicle because the speed will take the vehicle further. confidence assessment: 1
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09:11:56 It turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long. If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower. For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected. Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far. If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.
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RESPONSE --> When solving questions you need to consider all the elements that will effect the variables in the problem like air resistance. To be hones I am confused on the reasoning that this question requires to get the correct answers self critique assessment: 2
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09:22:44 `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length. Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?
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RESPONSE --> Yes based on the figures that I calculated a person of weight 125 lbs. would stretch the cord more than 7 feet beyond its initial unstretched length because as the weight increases from 100lb to 150lb the feet changes from 5 to 9. Which leads me to believe that if the weight is 125lbs which is between the 100lbs and 150lbs and 7 is between 5 and 9. confidence assessment: 1
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09:25:54 From 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft. A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).
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RESPONSE --> ok self critique assessment: 3
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09:33:03 `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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RESPONSE --> When the skater is pulled back 8 feet and released the skater will travel 40 feet which means she will be expected to travel twice as far as when she was pulled back only 4 feet. confidence assessment: 1
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09:34:26 06-03-2007 09:34:26 The distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far
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NOTES -------> The distance through which the force acts will be twice as great which alone doubles the distance.
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09:39:12 `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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RESPONSE --> To a moth seekign light from half a mile away the larger sphere will seem dimmer than the other sphere because the light will be brighter when the sphere is smaller. I think that the small sphere would appear to have the same brightness because the moth only see's 1 square inch so the size of the sphere makes no differnce in the question. confidence assessment: 1
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09:44:50 Both bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area. Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.
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RESPONSE --> I thought that the larger sphere would not be as bright as the smaller sphere because there was more area in the larger sphere which would cut down the brightness but I was wrong & I see the reasoning behind the answer, I guess I'm not thinking enough about the question. self critique assessment: 2
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10:14:07 `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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RESPONSE --> 1. Increasing the temperature of the ice by 20 degrees to reach its melting point 2. Increasing the temperatur of the water by 20 degrees after all the ice melted 3. Melting the ice at its melting point. To me it seems that the ice melts at the 40 degrees Celsius because that it the temperature that all the ice melts. confidence assessment: 0
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10:15:37 06-03-2007 10:15:37 Since the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius. The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.
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NOTES -------> It takes less energy to increase the temperature of ice than it does liquid water.
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10:26:59 `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you. Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much. If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point? How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?
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RESPONSE --> If the peaks of the approaching waves are each 6 inches high then a person would bob up and down 12 inches when you are at the center point. If you move almost to one side the peak and valleys will not meet and a person will not bob up and down so much,. confidence assessment: 0
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10:31:22 06-03-2007 10:31:22 end program
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NOTES -------> When calculating the amount someone needed to shift to be in the calmest water you need to divide the total feet by half to get the measurement for one side and the other half, then you have to divide the one side in half to get where someone would be in the calmest water.
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assignment #002 002. Volumes qa areas volumes misc 06-03-2007
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10:37:45 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> To solve this problem I multiply 3*5*7 = 105 confidence assessment: 1
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10:48:06 06-03-2007 10:48:06 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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NOTES -------> Volume of any regular solid is therefore V = A *h Where A is the area of the base and the h is the altitude. Volume of a rectangle is sometimes expressed as V = L*W*h, however the relationship V = A * h applies to a much broader class of objects than just rectangular solids, V = A * h is a more powerful idea than V = L* W*h and is more important.
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10:48:51 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> To solve this question I multiplied 48 by 2 to get the answer of 96. confidence assessment: 1
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10:49:54 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> I am getting the answers correct but I am confused about the m^2 and the m^3, I'm not sure why that is in the equation. self critique assessment: 1
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10:54:40 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> The answer I came up with is 12566.4 and I got this answer by taking pi*10^2*40. confidence assessment: 0
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10:57:20 06-03-2007 10:57:20 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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NOTES -------> V = A * h applies to uniform cylinders as well as to rectangular solids. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This the case for uniform cylinders and uniform prisms.
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11:01:41 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> The answer to this question that I got was 2356.2 and I got this answer by multiplying pi*5^2*30. confidence assessment: 1
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11:05:03 06-03-2007 11:05:03 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
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NOTES -------> When the cylinder is uniform it mean that its cross-sectional area is constant. The cross-sectional area A is the area of a circle so you figure the area of a circle which is A = pi*r^2 then you take that area and use the formula V = A * h to solve the equation
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11:06:13 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> I am not sure how to even solve this question. confidence assessment: 0
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11:10:33 06-03-2007 11:10:33 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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NOTES -------> Commonly we estimate the dimensions of a can of food in centimeters or inches but we can also use millimeters, feet, meters, and mile. Differnt cans will have different dimensions and the estimate will depend on the can being used. You can use the formula V = A * h where A is the area of the cross section. To find area use A = pi*r^2
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11:14:58 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> I used the formula V = A * h to work out this question. V = 50*60 = 3000 confidence assessment: 1
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11:24:42 06-03-2007 11:24:42 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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NOTES -------> For pyramids the formula V = A * h can't be used to solve the question because the thing doesn't have a constant cross - sectional area - from the base to apex the cross sections gets smaller and smaller. The volume of a pyramid is found by using the formula V = 1/3 * A * h
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11:26:46 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> I used the formula V = A * h to solve this question. V = 20 * 9 =180 confidence assessment: 1
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11:31:22 06-03-2007 11:31:22 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
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NOTES -------> The volume of a cone can't be found by using the formula V = A * h like the pyramid. The cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. Use the formula V = 1/3 A * h to find the volume of a cone. A is the area of the base and h is the altitude of the cone.
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11:34:57 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> The answer that I got was 268.1 and I used the formula V =4/3* pi* r^3 V = 4/3 * pi * 4^3 =268.1 confidence assessment: 1
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11:37:46 06-03-2007 11:37:46 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
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NOTES -------> I used the same formula to work the problem and I have rechecked my work and I don't see where I went wrong since our answers were differnent Formula to find the volume of a sphere is V = 4/3 pi r^3.
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11:49:02 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> I used the formula V = 4/3*pi*r^3 to get the answer 1.43 which I know is incorrect.
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11:55:42 06-03-2007 11:55:42 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
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NOTES -------> Formula for volume of a sphere is V = 4/3*pi*r^3 becareful when calculating the figures so you don't get questions wrong due to errors
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12:01:22 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> The principle of the base area times the height confidence assessment: 0
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12:03:08 06-03-2007 12:03:08 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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NOTES -------> The principal is that when the cross-section of an object is constant , its bolume is V = A * h where A is the cross- sectional area and h the altitude. Altitude is measure perpendicular to the cross section.
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12:11:19 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> I don't know confidence assessment: 0
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12:14:24 06-03-2007 12:14:24 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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NOTES -------> Pyramids and Cones The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h where A is the area of the base and h the altitude as measured perpedicular to the base.
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12:15:17 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> The formula for the volume of a sphere is V = 4/3*pi*r^3. confidence assessment: 3
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12:15:31 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> ok self critique assessment: 3
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12:19:07 06-03-2007 12:19:07 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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NOTES -------> I learned that the formula V = A * h applies to almost any solid except the pyramid and the cone. Radius is important when figuring out the volume of circles, cylinders, cone, and sphere.
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assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 06-03-2007
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12:33:34 `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 --> I solved this question by multiplying 3*4*6 = 72. confidence assessment: 1
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12:38:51 06-03-2007 12:38:51 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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NOTES -------> A rectangular solid has six faces top, bottom, front, back, left side, right side. The pairs top and bottom, right and left sides, and front and back have the same areas. Ex. 3,4,6 3 by 4 (12), 3 by 6 (18) 4 by 6 (24). 2(12)*2(18)*2(24) = 108
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12:47:49 `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 --> To solve this equation I multiplied 2*pi*5*12 = 376.9 rounded up to 377, this is surface area of the curved sides of a cylinder. To find the total surface area of the cylinder I took 2*pi*5 + 2*pi*5^2 = 377 + 157 = 534 confidence assessment: 1
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13:01:06 06-03-2007 13:01:06 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 -------> Circumference of this cylinder is 2*pi*r Area of curved side is A = circumfernce * altitude When a cylinder has a top and a bottom it is considered a closed cylinder. The total area of a closed cylinder is total area = area of sides + 2* area of the base
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13:03:33 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> I divided 3 into half to get the radius which is 1.5 Then I multiplied 4*pi*1.5^2 = 28.3 confidence assessment: 0
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13:05:49 06-03-2007 13:05:49 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 -------> Surface area of a sphere A = 4*pi*r^2
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13:08:35 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> The hypotenuse of this right triangle is14. I found this answer by adding 5+9=14 What the legs add up to equals the other side which is the hypotenuse. confidence assessment: 2
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13:11:23 06-03-2007 13:11:23 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 -------> I know what I did wrong on this one I didn't square the numbers like I should have. 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
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13:21:03 `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 --> I solved this question by using a^2+b^2=c^2 to solve this problem I squared 6 and 4 to get 36 and 16. Then I subtracted 16 from 36 to get 20. So the other side is 10 because 10 squared is 20 confidence assessment: 0
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13:22:37 If c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m.
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RESPONSE --> I know what I did wrong when I got to the answer 20 I should have gotten the square root of that number for the answer. self critique assessment: 2
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13:28:46 `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 --> I have no idea how to figure the density confidence assessment: 0
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13:38:03 06-03-2007 13:38:03 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 -------> Find the volume of the solid first by multiplying all 3 dimensions together. Ex. 4 cm*7cm*12cm = 336 cm^3 then to find the density we divide the number of grams by the number of cm^ When solids are said to be uniform and homogeneous it means that its all made of the same material which is uniformly distrbuted. For material that we don't know if its uniform were take the average density and divide it by the cm^3
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13:44:25 `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 --> I found the volume 4/3*pi*4^3 = 268 then I divided 3000/268 = 11.2 is the answer that I got. confidence assessment: 0
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13:49:26 06-03-2007 13:49:26 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 -------> Mass = density * volume I have no idea why I use the same formulas as you but I don't get the same answer as you and I don't understand, when I recheck myself I'm still getting it wrong?
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13:53:50 `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 --> To solve this question I took 4^3= 64 and 2^3=8 64+8=72 72/2=36 confidence assessment: 0
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13:59:53 The first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.
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RESPONSE --> The average density = total mass / total volume 4*6=24 & 2*10=20 24 + 20 = 44 grams the total mass Then divide 44 by 16 (6cm^3 + 10cm^3) = 2.75 grams / cm^3 self critique assessment: 2
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14:04:21 `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 --> The further I get into these question the more lost I seem to get. confidence assessment: 0
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14:05:35 06-03-2007 14:05:35 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 -------> Average density = total mass / total volume.
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14:10:09 06-03-2007 14:10:09 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 -------> To find the size of the oil slick you find the volume using the formula V = A * h and then you just enter your solution into the formula for mass which is density * volume
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14:12:14 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> We find the surface area of a cylinder by finding the circumference of the cylinder which is found by multiplying 2*pi*r confidence assessment: 0
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14:15:18 06-03-2007 14:15:18 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 -------> To find the surface area of a cylinder you use the formula 2*pi*r*h + 2*pi*r^2 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
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14:17:46 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The formula for the surface area of a sphere is 4*pi*r^2 confidence assessment: 2
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14:18:03 06-03-2007 14:18:03 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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NOTES ------->
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14:22:42 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is the relation between mass and volume of an object. confidence assessment: 0
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14:23:51 06-03-2007 14:23:51 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 -------> If an object is uniform and homogneous then its dentsity is constant and we can speak of its 'density' as opposed to its 'average density'
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14:26:52 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Since we multiply density and volume to get the mass of an object then we can divide the mass by the density and should get what the volume should do. confidence assessment: 0
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14:27:06 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> ok self critique assessment: 3
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14:30:14 `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 organized my knowledge by saving the formulas as notes for future reference. This assigment made me realize I need to review the information covered. confidence assessment: 2
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