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10:22:39 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> The two poblems are handled differently...since the brackets dictate which operations are performed first. The problems follow the Multiplication / Division / Addition / Subtraction rule in that particular order where anything in which operations inside brackets would be performed first. Thus the operations in brackets shown in the second problem would be performed first. The first problem the first operation would be the division.
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10:23:10 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction.
It makes a big difference whether you subtract the 2 from the 2 or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract: 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.......!!!!!!!!...................................
RESPONSE --> Correct
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10:31:39 `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2.
Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.......!!!!!!!!...................................
RESPONSE --> Same response as the previous problem as to the operations in the parenthasis are performed first. In this case the the parenthesis place the +4 into the exponentation as well see below:
2^x+4 = 2^(2)+4 = 4+4 = 8 2^(x+4) = 2^(2+4) = 2^6 = 64 major difference in the way the problem is worked..................................................
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10:31:55 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4.
2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.......!!!!!!!!...................................
RESPONSE --> Correct
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11:07:32 `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
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RESPONSE --> Numerator = 3 Denominator = [(2x-5)^2*3x+1]
When evaluated the expression "after my work" comes out to 3/-11-14.................................................
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11:10:53 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3.
If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.......!!!!!!!!...................................
RESPONSE --> My error in this problem was that I made is much much more complecated than needed to be and made multiple errors in the process. In my work I actually factored out the problem and tried to simplify it before using subsituting 2 in as x....I have reworked the problem just substituting 2 in as x and working the problem. Much easier and I arrived at the same answer as you have listed to the side
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11:17:46 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> (x-5)^2x-1+3/x-2 ([4]-5)^2(4)-1+3/4-2 = 4-1+3/4-2 = 3+3/4-2 = Answer 1.75
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11:20:21 We get
(4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication. Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4). Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.......!!!!!!!!...................................
RESPONSE --> I did not leave in fraction form...but my answer can out the same.
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τzo|{|q Student Name: assignment #002 002. Describing Graphs
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11:41:10 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor.
Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.......!!!!!!!!...................................
RESPONSE --> Complete -- I will email assignment as requested
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11:43:41 The graph goes through the x axis when y = 0 and through the y axis when x = 0.
The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.......!!!!!!!!...................................
RESPONSE --> OK.
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11:48:54 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> For each unit of change in the value of y in the positive direction the graph changes value of a total of 3 positive units in the x direction plus one positive unit in the y direction.
For each unit of change in the value of y in the equation the exact opposite happens.................................................
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11:49:36 The graph forms a straight line with no change in steepness.
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RESPONSE --> Sorry...the graph does for a straigh consistent line
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11:50:22 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> 3/1
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11:51:00 Between any two points of the graph rise / run = 3.
For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.......!!!!!!!!...................................
RESPONSE --> OK
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11:55:55 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?......!!!!!!!!...................................
RESPONSE --> The graph is increasing at an increasing rate
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11:57:47 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right.
The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate.......!!!!!!!!...................................
RESPONSE --> OK
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12:02:47 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0.
Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?......!!!!!!!!...................................
RESPONSE --> The line is decreasing; the steepness of the graph is decreasing at a decreasing rate
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12:03:03 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.
Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.......!!!!!!!!...................................
RESPONSE --> OK
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12:07:51 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?......!!!!!!!!...................................
RESPONSE --> It appears that the graph is increasing at a decreasing rate
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12:08:02 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing.
The graph would be increasing at a decreasing rate.{}{} If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. {}{}If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.......!!!!!!!!...................................
RESPONSE --> OK
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12:13:01 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3.
Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?......!!!!!!!!...................................
RESPONSE --> It appears the graph is decreasing at a decreasing rate
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12:13:08 ** From basic algebra recall that a^(-b) = 1 / (a^b).
So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **......!!!!!!!!...................................
RESPONSE --> OK
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12:14:36 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.
If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?......!!!!!!!!...................................
RESPONSE --> Increasing at a increasing rate
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12:14:46 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> OK
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ַmᳱdlC{ÿ́ Student Name: assignment #001
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12:20:50 `q001. It will be very important in this course for your instructor to see and understand the process of visualization and reasoning you use when you solve problems. This exercise is designed to give you a first experience with these ideas, and your instructor a first look at your work.
Answer the following questions and explain in commonsense terms why your answer makes sense.......!!!!!!!!...................................
RESPONSE -->
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12:22:22 For each question draw a picture to make sense out of the situation, and include a description of the picture.
Samples Sample question and response Question: If a bundle of shingles covers 30 square feet, how many bundles are required to cover a 600 square foot roof? Response: We might draw a picture of a rectangle representing the area, dividing the rectangle into a number of smaller rectangles each representing the area covered by a single bundle. This makes it clear that we are dividing the roof area into 1-bundle areas, and makes it clear why we are going to have to divide. Reasoning this problem out in words, we can say that a single bundle would cover 30 square feet. Two bundles would cover 60 square feet. Three bundles would cover 90 square feet. We could continue in this manner until we reach 600 square feet. However, this would be cumbersome. It is more efficient to use the ideas of multiplication and division. We imagine grouping the 600 square feet into 30 square foot patches. There will be 600 / 30 patches and each will require exactly one bundle. We therefore require 600 / 30 bundles = 20 bundles. {}Your responses might not be as clear as the above, though they might be even more clear. I won't be looking for perfection, though I wouldn't object to it, but for a first effort at visualizing a situation and communicating a reasoning process. This is not something you are used to doing and it might take a few attempts before you can achieve good results, but you will get better every time you try. {}You might be unsure of what to do on a specific question. In such a case specific questions and expressions of confusion are also acceptable responses. Such a response must include your attempts to come up with a picture and reason out an explanation. For example your response might be Sample expression of confusion: I've drawn a picture of a pile of bundles and a roof but I'm not sure how to connect the two. I tried multiplying the number of bundles by the square feet of the roof but I got 18,000, and I know it won't take 18,000 bundles to cover the roof. How do you put the area covered by a bundle together with the roof area to get the number of bundles required? A poor response would be something like 'I don't know how to do #17'. This response reveals nothing of your attempt to understand the question and the situation. Nor does it ask a specific question. Incidentally, you might be tempted to quote rules or formulas about rates and velocities in answering these questions. Don't. This exercise isn't about being able to memorize rules and quote them. It is about expanding your ability to visualize, reason and communicate.......!!!!!!!!...................................
RESPONSE --> OK
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12:23:34 In your own words briefly summarize the instructions and the intent of this exercise.
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RESPONSE --> The intent of the excecise is to refelect to the instructor on how the student can take a written problem and describe and solve such problem in written and visual form
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12:29:58 `q001. If you earn 50 dollars in 5 hours, at what average rate are you earning money, in dollars per hour?
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RESPONSE --> If I am earning 50 dollars in 5 hours, I know that I am earning 1/5 of 50 dollars in each hour that I work. In other words I can take the total amount of money earned over the 5 hour period and divide it by the number of hours worked and I would get the total amount of money earned per hour.
--See visual--.................................................
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12:36:12 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
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RESPONSE --> I know that it take 6 hours to travel 300 miles...I want to know on the average how long it took me to travel only one mile. I would simply divide the 300 miles by 6 hours which would give me the total miles traveled per hour: Answer 50
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12:46:57 `q002. If a ball rolling down a grooved track travels 40 centimeters in 5 seconds, at what average rate is the ball moving, in centimeters per second?
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RESPONSE --> We know that the ball has traveled 40 cm in 5 seconds...If we divide 40 cm total distance traveled by 5 seconds the total time traveled...we would get an average rate of 8cm/second.
We realize to average 40cm of distance over 5 seconds. The ball would need to be moving an average of 8 cm for each second of time passed..................................................
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12:51:56 The preceding three questions illustrate the concept of a rate. In each case, to find the rate we divided the change in some quantity (the number of dollars or the distance, in these examples) by the time required for the change (the number of hours or seconds, in these examples). Explain in your own words what is meant by the idea of a rate.
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RESPONSE --> Rate is the relationship of quantities such as dollars per hour or traveled distance per hour. The change of rate is the adjustment of one or more of the variables to account for the change in its value. For example if you make $1 per hour your would need to adjust the rate to reflect the total amount earned in 2 hours. Same with distance time. If you travel 1 mi / hr on the average you can use this rate to reflect what lenght of time it takes to travel 5 miles..or you can use this rate to calculate how far you have traveled in 12 hours.
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12:55:30 `q003. If you are earning money at the average rate of 15 dollars per hour, how much do you earn in 6 hours?
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RESPONSE --> If you know that for every hour worked you recieve $15 dollars then you can calulate that having worked 6 hours you will earn a total of $90.00.
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12:57:08 If you are traveling at an average rate of 60 miles per hour, how far do you travel in 9 hours?
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RESPONSE --> If we know that we are traveling 60 miles for each hour traveled...and we know that we have traveled 9 hours we can multiply the total hours traveled by 60 (total distance traveled in 1 hour)
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12:58:42 `q004. If a ball travels at and average rate of 13 centimeters per second, how far does it travel in 3 seconds?
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RESPONSE --> If we know that the ball travels 13 cm per second and we have traveled 3 seconds, then we can multiply the total time traveled by the distance traveled in 1 sec.
Answer: 39 cm.................................................
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12:59:17 In the preceding three exercises you turned the concept of a rate around. You were given the rate and the change in the clock time, and you calculated the change in the quantity. Explain in your own words how this increases your understanding of the concept of a rate.
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RESPONSE --> This gives a better understanding of the rate of change in values compared to each variable
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13:00:54 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
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RESPONSE --> We know we need to earn $100, but we can only earn it in a rate of $4.00 per hour. If you take the total amount needed to earn $100 and divide it by the rate of money we can earn in 1 hour....you will realize that it will take 25 hours of work to earn the required $100
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13:02:36 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
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RESPONSE --> We know we need to travel 500 miles, but the average rate we can travel per hour is only 25 miles. If we divide the total distance we need to travel by the rate at which we travel at...we get a total time of 20 hours required to travel 500 miles at an average rate of 25 mi /hr
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13:04:26 `q006. How long does it take a rolling ball to travel 80 centimeters at an average rate of 16 centimeters per second?
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RESPONSE --> We want to know how long it takes the ball to travel 80cm when the average rate of travel is 16cm per second. If we divide the total length needed to travel 80cm by the average rate of 16cm per second we know that it will take 5 seconds for the ball to travel 80cm
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13:04:36 In the preceding three exercises you again expanded your concept of the idea of a rate. Explain how these problems illustrate the concept of a rate.
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RESPONSE --> OK
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꾔 ʟ̌ma Student Name: assignment #001 001. Rates
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13:06:22 `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'.......!!!!!!!!...................................
RESPONSE --> OK
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13:06:31 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
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13:07:21 `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?......!!!!!!!!...................................
RESPONSE --> $10/hr
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13:07:43 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.......!!!!!!!!...................................
RESPONSE --> OK
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13:08:43 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> $5000.00 per month (Would certainly love to have this to calculate as my own personal paycheck)
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13:08:53 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.......!!!!!!!!...................................
RESPONSE --> OK
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13:10:54 `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 --> An average of $5000.00 would be more appropriate...since the revenue of the company would more than likely not be exactly $5000.00 per month. The actual amount would vary from month to month thus the average would be more correct
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13:11:05 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.......!!!!!!!!...................................
RESPONSE --> OK
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13:13:33 `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 --> Would be traveling an average of 50 miles per hour. The average is used instead of the plain rate simple because the actual rate of travel at any point in the trip more than likely will change. This trip could include a stop or two along the way of which is also "averaged" into the total average time/distance of the trip.
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13:13:42 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.......!!!!!!!!...................................
RESPONSE --> OK
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13:15:05 `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 --> will be consuming gas at a rate of 20 MPG or miles per gallon
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13:15:39 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.......!!!!!!!!...................................
RESPONSE --> I need to pay more attention to the directions..
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13:19:05 `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 --> an average and average rate speak in different terms. An average rate deals with a unit in respect to time, distance, etc. It is a ratio of seperated quantites. Average time traveled / Average distance traveled, etc. We are taking the
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13:19:20 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.......!!!!!!!!...................................
RESPONSE --> OK
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13:30:51 `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 --> Group 1: 10 pushups per day for a year Group 2: 50 pushups per day for a year
Both groups averaged 30 pushups per day Both groups averaged 154.5 lifting strenght year end 30 pushups per day = 10950 pushups per year Both groups averaged .014 lifting strenght per each push up. Group 1: Averaged .04 lifting strenght per each push up Group 2: Averaged .008 lifting strenght per each push up.................................................
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13:33:24 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.......!!!!!!!!...................................
RESPONSE --> Again, I over complicated the calculations...Instead of looking for a pattern rate to use in calculation. I did also misunderstand the value of the lifting strenght....I calculated that as a year end increase rather than the comparison to the two groups.
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13:35:29 `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 --> Group two recieved 17 more pounds of lifting strenght with the additional 20 pounds of weight.
Answer .85.................................................
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13:35:49 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.......!!!!!!!!...................................
RESPONSE --> OK -- Kept is much simpler...:-)
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13:37:30 `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 --> 10m / second
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13:37:36 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.......!!!!!!!!...................................
RESPONSE --> OK
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13:46:29 `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 --> Assuming that you are asking for a rough calculation of how long it takes the runner to cover the distance between the 200-meter mark and the 300-meter mark...
Based on the 100-meter and 200-meter marks...it seems that for every 100 m the runner travels in the inital stages of the race the runner tires. My best estimate of how long it takes to cover the next 100 meter distance is 8 meters / second.................................................
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13:47:34 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.......!!!!!!!!...................................
RESPONSE --> I do realize that averaging the first two 100m would be a much better estimate..
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13:48:24 `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 --> Because we are talking about two seperate average rates between two seperate distance marks.
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13:48:38 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.......!!!!!!!!...................................
RESPONSE --> ok
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}i |rÒlFH} Student Name: assignment #001
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14:22:16 `q001. The frequency of a pendulum is how frequently it oscillates back and forth. A very short pendulum oscillates much more quickly than a very long pendulum.
A cycle is a complete oscillation, from one extreme point to the other and back. Frequency is usually measured in cycles/second. However, it could be measured in cycles/minute, cycles/millisecond, cycles/year, or cycles/(time unit), where (time unit) is any unit of time. In this experiment you will observe and measure frequency in cycles/minute. If you hang your arm loosely at your side and nudge it a bit, it should oscillate back and forth a few times. Try it. Estimate how many complete oscillations it will undergo in a minute if you keep nudging it slightly to keep it moving.......!!!!!!!!...................................
RESPONSE --> Aprox 48
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14:22:32 If you managed to keep your arm relaxed it probably took a bit over a second to complete one back-and-forth oscillation. For most people a relaxed arm will oscillate about 40 to 50 times in a minute.
If your arm is tense, then you probably had to force the oscillations in you might have ended up with anything from about 30 to 100 times a minute.......!!!!!!!!...................................
RESPONSE --> OK
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14:25:14 `q002. In this activity you will:
1. Observe the frequency f of a pendulum vs. its length L. 2. Make a table of f vs. L. 3. Make a graph of f vs. L. Note the following conventions for determining which variable is independent and which dependent, and for placement of the dependent and independent variables on tables and graphs. f depends on L. We control L by holding the pendulum string at different lengths and observe its effect on f, so 1. f is the dependent variable, L the independent. 2. When we make a table, the independent variable goes in the first column, the dependent in the second. 3. When we make a graph, the independent variable goes horizontally across the page, and the dependent variable up and down (dependent is vertical vs. independent, which is horizontal). Note how this relates to the traditional way of graphing y vs. x, with y vertical and x horizontal. Summarize these conventions in your own words.......!!!!!!!!...................................
RESPONSE --> I will create a table in two colomns to include f and L; f is the dependent variable and will follow the y axis, L is the independent variable and will follow the x azis
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14:26:51 Predictions
Imagine a rock or some other mass hanging by a string over the rail of a high deck. As the rock swings back and forth, you gradually let more string over the rail, lowering the mass. Will the swings take longer and longer, or will they become more and more frequent? If it requires 2 seconds for the rock to swing back and forth from a certain length of string, how long do you think it would take if the string was twice as long? What if the string was half as long?......!!!!!!!!...................................
RESPONSE --> Lowering the mass would in my opinion the swings will take longer and longer
If at 2 seconds: Twice as long 4 seconds half as long 1 second.................................................
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14:30:11 `q003. Sketch a graph of the time required for a swing vs. pendulum length, as you would predict it. Describe your graph in some detail, using the conventions outlined in the 'Describing Graphs' exercise.
As with all graphs, be careful to use consistent units on each axis. If the marks on an axis are equally spaced, then they should represent the same change in the quantity represented by that axis. A common error would be, if for example the numbers 5, 10 and 20 were to be represented, would be to make three equally spaced marks representing the numbers 5, 10 and 20. The spacing between 5 and 10 should be half the spacing between 5 and 20. If you're not careful about your spacing, you will distort the shape of your graph.......!!!!!!!!...................................
RESPONSE --> OK
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14:30:40 If you said, as many people do, that a string twice as long to would imply a 4 second period (the period is the time per swing) and that a string half as long would imply a 1 second period, then you probably are assuming a linear relationship between period and length. In this case your graph should have been a straight line, a line which is decreasing at a constant rate.
If you said that a doubled length implies less than a doubled period, then to be consistent you probably said that half the length implies more than half the period. In this case your graph would be increasing but at a decreasing rate. Other responses are possible. The experiment will tell you whether your prediction was correct.......!!!!!!!!...................................
RESPONSE --> OK
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14:36:04 `q004. If a pendulum of a certain length swings back and forth 35 times per minute, how many times per minute do you think it would take if the string was twice as long? What if the string was half as long?
Sketch a graph of the number of swings per minute vs. pendulum length, as you would predict it. Describe your graph. Be sure to describe how your graph behaves as pendulum length approaches zero.......!!!!!!!!...................................
RESPONSE --> Graph is linear and increasing at a constant rate. As the pendulum lenght approaches zero so does f.
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14:36:23 The most common responses that the pendulum would swing back and forth 70 times per minute if the string was half is long, and 17.5 times per minute if the pendulum was twice as long. This would be consistent with the most common answer to the preceding question, which resulted in an increasing linear graph. The present graph would be decreasing, as we would expect, and would still be linear.
A prediction that half the length would imply less than 70 cycles per minute would be consistent with the idea that doubling the length will imply more than 17.5 cycles per minute. This would result in a graph that decreases but at a decreasing rate.......!!!!!!!!...................................
RESPONSE -->
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14:36:31 `q005. How are your answers to the first two questions related to your answers to the last two? Are they consistent?
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RESPONSE --> YES
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14:39:00 The Experiment
Frequency is measured in cycles/time units. Here we will measure frequency in cycles/minute, because it's easy to count the cycles in a minute. First make the pendulum: Tie a light string around a rock, a potato, or any relatively small, dense object. The length of the pendulum is measured from the fixed point of the swing to the center of gravity of the object. Describe how you constructed your pendulum.......!!!!!!!!...................................
RESPONSE --> The object is a small rock attached to a string
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14:40:11 `q006. Get the 'feel' of the pendulum.
Feel how quickly a pendulum 1 foot long swings back and forth. Don't time anything yet; just get the feel of the thing. Then get the feel of the rhythm of a two foot pendulum, then a three foot pendulum, then a4 foot pendulum. Try walking to the rhythm of each, stepping (either one or two steps to the cycle) the same distance on each 'beat' of the pendulum. Does the speed of your walk change by the same amount each time the length of the pendulum increases by a foot?......!!!!!!!!...................................
RESPONSE --> Yes...my pace tends to get slow as the pendulum length increases.
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14:41:25 What would a graph of your speed vs. time look like? Would it be a straight line, would it curve upward, or would it go upward and tend to level off?
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RESPONSE --> It seems it would be a straight line
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15:01:30 `q007. Observe the number of cycles in a minute for different lengths:
Measure different lengths, from about a foot to the longest pendululum you can easily manage. Increase the length by the same amount each time, and time the pendulum for a minute at each length. Choose your length increment so you will end up with about six observations (i.e., so you will have six different lengths). Don't let the pendulum swing too far (no more than about 10 degrees from vertical). The frequency f, in cycles/minute, is the number of complete cycles in the minute. Write down in a table the frequency f and length L for each length. Let the first column be the length, the second the frequency. Label this table Data Set 1, and include the table in your response here.......!!!!!!!!...................................
RESPONSE --> Data Set 1
L f .5ft 70 1ft 51 1.5ft 42 2ft 37 2.5ft 33 3ft 30.................................................
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15:02:40 Analyzing The Experiment
Look at your data for Data Table 1 and see how changes in frequency are related to changes in length: Look at the numbers on the table. Do the frequencies change regularly with the lengths, or to they change faster and faster, or slower and slower?......!!!!!!!!...................................
RESPONSE --> as the lenght of the pendulum string get smaller the frequency chages faster and faster
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15:05:31 `q008. As you look at the numbers, try to visualize the graph.
Sketch a graph of the general shape of your f vs. L data. Don't mark off a scale, don't plot points, just sketch the basic shape, from your examination of the numbers.......!!!!!!!!...................................
RESPONSE --> ok
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15:07:21 Describe the graph.
How does the graph reflect the behavior of the numbers on the table? What is it about the graph that is related to the 'feel' of the walking exercise?......!!!!!!!!...................................
RESPONSE --> As the pendulum lenght gets longer the rate is decreasing at a decreasing rate.
The walking exercise gave a good feel for the rate of the pendelum at both a short lengh of string and long length of string.................................................
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15:07:49 As the length increases, your table will show that the frequency decreases. That is, for a longer pendulum you will count fewer complete cycles in a minute.
Your table will also show that equal increases in the length of the pendulum result in smaller and smaller decreases in the frequency. For example, between a 1-foot pendulum and a 2-foot pendulum the frequency changes from about 55 cycles per minute to about 40 cycles per minute, the decrease of about 15 cycles per minute, while changing from a four-foot pendulum to a five-foot pendulum the decrease would be from about 28 cycles per minute to about 25 cycles per minute, a decrease of only about 3 cycles per minute. Note that the numbers given here are very approximate, so if your results differ by a few cycles per minute it is no cause for alarm. This behavior will cause the graph to decrease at a decreasing rate. With the walking exercise you should have noticed that each increase in length resulted in a decrease in walking speed, but that the decrease was less for the longer pendulums than for the shorter.......!!!!!!!!...................................
RESPONSE --> OK
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15:11:57 `q009. Sketch the graph by marking a scale and plotting points and compare with your rough sketch. How does the general shape of your graph compare with the shape of the rough sketch?
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RESPONSE --> Fairly close....
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15:30:16 Consider the period of the pendulum vs. length:
The period of a pendulum is the time required for 1 complete cycle. That is, if the pendulum requires 2 seconds to complete a cycle, the period is 2 seconds. Before doing any calculations to find the period, recall your direct experience of the pendulum. Does the period increase or decrease with length? What do you think a graph of the period vs. length would look like? Sketch a rough graph of period vs. length, and describe it in words.......!!!!!!!!...................................
RESPONSE --> The period will increase with lenght
I placed the period in the y axis and the Lenght in the x axis. The graph is increasing and appears to be at a decreasing rate.................................................
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15:32:18 `q010. From your data, figure out the period associated with each length. Make a table of period vs. length, label it clearly, and include the table in your response here.
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RESPONSE --> L Period .5 ft .85 s 1 ft 1.17 s 1.5 ft 1.42 s 2 ft 1.62 s 2.5 ft 1.81 s 3 ft 2 s
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15:38:58 The period is the number of seconds required for cycle. For example, if there are 30 cycles in a minute, then it takes 2 seconds for each cycle and the period is 2 seconds. If there are 40 cycles in a minute, 40 cycles require 60 seconds and a single cycle takes 60/40 seconds = 1.5 seconds.
If you miscalculated your periods, recalculate them and show the corrected table. Look at the numbers, asking yourself the same sort of questions as before. Visualize a graph of the numbers in this table. Then draw the graph. Can you look at the graph in such a way as to invoke the 'feel' of the things you have observed? Describe how the graph you made from the table is like, and how it is different than the graph you sketched before you made the table.......!!!!!!!!...................................
RESPONSE --> The graph that I originally made estimating how I thought a pendulum cycle would change in respect to the lenght of string was in a linear distance.
After the experiment I have learned that as the length of rope increases for the pendulum. The cycle time of the pendulum gets increasingly greater with each unit of length.................................................
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15:49:30 `q011. Let T stand for the period of the pendulum (the time required for a single cycle). You have a table of T vs. L. Now create a table for log(T) vs. log(L):
Use the log key on your calculator to obtain the required values. For example, if the period T is 3.1, log(T) will be approximately .5. For every T and every L on your table, find log(T) and log(L) and place these values in your new table. Show your table here.......!!!!!!!!...................................
RESPONSE --> L T -.301 -.07 0 .068 .176 .152 .301 .209 .397 .257 .477 .301
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15:55:01 Next plot your new table, obtaining a graph of y = log(T) vs. x = log(L).
If you have taken your data carefully, and if you have used your calculator and plotted your points correctly, your graph should form a straight line. If not, go back and fix anything that needs to be fixed and replot it. Now using a straightedge, draw a straight line through your graph points and extend this line to the vertical (y) axis; also extend it a little ways past your last data point. If your line doesn't exactly fit every point, make it come as close as possible, on the average, to your graph points. Describe your graph and how well the straight line fits it.......!!!!!!!!...................................
RESPONSE --> My line is almost fits within a straigh line...If an exact graphing program were used I feel that it would reflect a straight line. Estimating some of the coordinates on the graph took away from accuracy.
The graph is increasing at a constant rate.................................................
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16:06:43 `q012. By measuring the rise and the run between the left and right endpoints of your line (not between two points from your table), determine the slope of the line. Call this value p.
Determine the y-intercept of your straight line (i.e., the value of log(T) where your line meets the vertical axis). Call this value y0. Give your values of p and y0.......!!!!!!!!...................................
RESPONSE --> p=.371/.778 p = .476
y0 = .068.................................................
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16:11:24 Your pendulum model will be the function T = A * L^p, where p is the slope you determined in a recent step and A = 10^y0. Use your calculator to determine 10^y0 for your value of y0. What do you get for your value of A?
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RESPONSE --> A=10^.068 = 1.1694
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16:12:22 q013. Your function model is T = A * L^p. Substitute the values of A and p into this form. What is your function?
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RESPONSE --> T = 1.1694(L^.476)
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16:19:40 Compare your function T = A * L^p with your data, using the values you have found for A and p, as follows:
For every pendulum length L, calculate the period Tpred predicted by this function. That is, plug in each length L and get your predicted value of T, then call this value Tpred. What values do you get for Tpred?......!!!!!!!!...................................
RESPONSE --> L Tpred .5 .839 1 1.169 1.5 1.418 2 1.626 2.5 1.809 3 1.973
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16:44:28 `q014. For each pendulum length, calculate the difference T - Tpred.
Express each of these differences as a percent of Tpred. Give a table with columns for L, T, Tpred, T-Tpred, and percent difference. What is your maximum percent difference, and how nearly do you conclude that your function predicts the actual behavior of the pendulum? Does the function do a good job of predicting the period of your pendulum?......!!!!!!!!...................................
RESPONSE --> L T Tpred Tpred-T % .5 .85 .839 .01 1.2% 1 1.17 1.169 .001 0% 1.5 1.42 1.418 .002 .001% 2 1.62 1.626 .006 .003% 2.5 1.81 1.809 .001 .001% 3 2 1.973 .02 .01%
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16:48:09 The 'correct' function for T vs. L is one of the following two equivalant functions:
If L was measured in feet and T in seconds, the ideal function would be about T = 1.1 L^.5. If L was measured in centimeters and T in seconds, then the ideal function would be about T = .2 L^.5. How close was your function to the ideal function? Figure out the periods corresponding to the lengths you observed. How close are these 'ideal' periods to the periods you observed?......!!!!!!!!...................................
RESPONSE --> fairly close....but off within reason...If this is the "idea" case I am very happy with the results of the experiment
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