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15:21:44 `q001. In this experiment you will construct a pendulum using a thread and one of the washers that came with your initial pack of CDs. If you have to use something heavier than a thread you will need to report what the string was made of and how thick the string is.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.
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RESPONSE --> 40 TO 50 OSCILLATATIONS/MINUTE
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15:22:11 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.
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RESPONSE --> OK
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16:19:30 `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). This corresponds to the traditional way of graphing y vs. x, with y vertical and x horizontal. Summarize these conventions in your own words. And do note them since you will be using these conventions later in this experiment.
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RESPONSE --> MAKE A TABLE WITH A FIRST COLUMN OF L TO BE THE DEPENDANT FOR THE NEXT COLUMN OF F. THEN MAKE A GRAPH WITH L AS THE X-AXIS AND F AS THE Y- AXIS.
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16:50:57 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?
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RESPONSE --> IF THE STRING HAD LITTLE SLACK, THEN, THE BALL WOULD MOVE FASTER. IF THERE WAS A LONGER STRING, THE BALL WOULD MOVE SLOWER. IF THERE IS 2 SECONDS FOR THE FIRST LEGNTH, THEN IT WOULD BE 4 SECONDS FOR THE LONGER LEGNTH.
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16:53:15 `q003. A pendulum of length 1 foot will require about 1 second to complete 1 oscillation (from one extreme point in its motion to the other and back). How many oscillations would you therefore expect to count in a minute?
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RESPONSE --> 60= SECONDS PER MINUTE 1= OSCILLATION PER SECOND I * 60 = 60 SO 60 OSCILLATIONS PER MINUTE
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16:53:29 There are 60 seconds in a minute. If the pendulum completes one oscillation every second then you will count 60 complete oscillations in a minute. Note that the actual time isn't exactly one second, and you will not count exacly 60 complete oscillations in a minute, but when you make your count you will be fairly close to 60.
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RESPONSE --> OK
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16:59:53 `q004. Suppose for the moment that a pendulum of length 1 foot does in fact make exactly 60 oscillations in a minute, as it would if Earth's density and/or radius were just a little different. Then based on your experience of pendulums, how many oscillations do you think a pendulum of length 2 feet would make in a minute? How many oscillations do you think would be made in a minute by pendulums of length 3 feet and 4 feet?
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RESPONSE --> 1 FT = 60 OSCILLATIONS :: 2 FT = 30 0SCILLATIONS NOT SURE...
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17:01:19 You should have specified three estimates, one each for pendulum lengths 2, 3 and 4 feet. If you didn't, do so now.
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RESPONSE --> 2 FT = 120 SECS 3 FT = 210 SECS 4 FT = 408 SECS
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17:13:53 `q005. Suppose that a pendulum of a certain length made 120 oscillations in a minute. Then how much time, in seconds or fractions of a second, would be required for a single oscillation? If a pendulum of a certain length required 1.5 seconds for a complete oscillation then how many oscillations would it complete in a minute? Three different pendulums complete 20, 30 and 40 cycles, respectively, in a minute. How many seconds are required for a single oscillation of each pendulum?
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RESPONSE --> ,5 SECONDS FOR A SINGLE OSCILLATION IN A MINUTE FROM 120 OSCILLATIONS PER MINUTE. 60 / 1.5 = 40 SO 40 OSCILLATIONS FOR A COMPLETE MINUTE FOR THE PENDULUM THAT REQUIRED 1.5 SECONDS FOR ONE OSCILLATION . 60/20 = 3 SECONDS PER OSCILLATION 60/30 = 2 SECONDS PER OSCILLATION
60/40 = 1.5 SECONDS PER OSCILLATION.................................................
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17:14:20 If a pendulum completes 120 oscillations in 60 seconds then it completes two oscillations in a second. An oscillation therefore requires 1/2 second, or .5 second. If a pendulum requires 1.5 seconds to complete an oscillation, then in 60 seconds it will complete (60 seconds) / (1.5 second / cycle) = 40 cycles. Note that since it takes more than one second to complete an cycle, fewer than 60 cycles will be completed in a minute. This is your cue to divide rather than multiply 60 by 1.5. If a pendulum completes 20 cycles in 60 seconds, then it should be clear that 3 seconds are required for each cycle. The formal calculation is 60 seconds / (20 cycles) = 3 seconds / cycle. It should also be clear that 30 cycles in 60 seconds implies 2 seconds for each cycle. Formally, 60 seconds / (30 cycles) = 2 seconds/cycle. The answer for 40 cycles in a minute might not be as clear, so the formal calculation is useful. Just as in the previous examples we need to divide the number of seconds by the number of cycles. This time we obtain 60 seconds / (40 cycles) = 1.5 seconds/cycle. Each cycle takes 1.5 seconds.
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RESPONSE --> I AGREE ABOUT THIS.
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17:22:20 `q006. According to your previous predictions, will it require more or less than 1 second for a 2-foot pendulum to complete a cycle? Will the time required for a cycle increase or decrease as the pendulum gets longer? According to your predictions of the number of cycles expected in a minute, how long would it require for a pendulum of each length 2, 3 and 4 feet to complete a single oscillation?
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RESPONSE --> IT WILL REQUIRE MORE TIME FOR A 2 FT PENDULUM TO COMPLETE A CYCLE. THIS TIME WILL INCREASE AS THE PENDULUM GETS LONGER. 1 FT = 60 OSCILLATION PER MINUTE 2 FT = 30 OSCILLATIONS PER MINUTE=2 SND 3 FT = 15 OSCILLATIONS PER MINUTE=4 SND 4 FT = 7.5 OSCILLATIONS PER MINUTE=8 SND BECAUSE 60SEC/30 OSC = 2 SECONDS PER OSCILLATION AND THE REST ARE IN THE SAME MANNER.
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17:23:56 You should in each case have divided 60 seconds by the predicted number of oscillations, as was done in the preceding problem. If you did not do so, do so now and enter your corrected response.
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RESPONSE --> 60/30 = 2 60/15 = 4 60/7.5= 8
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17:34:11 `q007. Now using the conventions outlined earlier, sketch a graph of the time required per cycle vs. the length of the pendulum. Be careful about which axis you use for the time per cycle and which axis for the length of the pendulum. Using the conventions outlined in the 'Describing Graphs' exercise (i.e., using the language 'increasing at an increasing rate', 'increasing at a decreasing rate' etc.) give a description of your graph.
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RESPONSE --> WELL THE GRAPH HAS AN X - AXIS OF - 90 TO POSITIVE 140, WITH THE Y- AXIS HAVING NEGATIVE 150 TO POSITIVE 150.
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17:40:08 If you assumed, 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, as many people also do, 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.
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RESPONSE --> OK ASSUMING THAT I HAVE A 20 FT STRING THEN I'M GUESSING ABOUT 80 SWING PERR MINUTE.
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17:44:50 `q008. Construct a pendulum 1 foot long (that's 30.5 cm in case you don't have a ruler that measures feet) by tying a light string or thread to the washer in your initial materials. Holding the string as nearly stationary as possible, with a small nudge allow the pendulum to swing a few inches, no more, between its extremes observe the number of cycles in a minute. Get the most accurate count you can, estimating if possible even fractions of a cycle. Make at least three one-minute counts to see if your results are consistent. Give your results and state just how accurately you think you have counted (e.g., plus or minus 2 cycles per minute, or plus or minus 1/2 cycle per minute).
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RESPONSE --> THERE ARE AT LEAST 60 OSCILLATIONS PER MINUTE. THE SECOND TIME ABOUT THE SAME AND THE THIRD TIME ABOUT THE SAME.
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17:45:05 The accurate number of cycles in a minute is easily calculated from a certain formula. It turns out that the number is very close to 60 / sqrt(1.22). Calculate this number and compare your result to this result. How close did you come?
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RESPONSE --> VERY CLOSE.
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17:45:29 `q009. How does your 1-minute count of an actual pendulum change your estimates of the counts for lengths of 2, 3 and 4 feet? What are your revised estimates?
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RESPONSE --> THEY REALLY DIDN'T CHANGE. THEY STAYED ABOUT THE SAME.
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17:45:47 Your count should have been around the ideal value of 54, rounded to the nearest whole number. This is 54 / 60 = .9 of the 60 counts we assumed originally. It would therefore be reasonable to multiply estimates for 2, 3 and 4 feet by .9 to take account of the observations you have now made.
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RESPONSE --> OK
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17:47:15 `q010. Observe the number of cycles in a minute for different lengths: Measure different lengths, from 1 foot to 4 feet in increments of 1 foot. Increase the length by the same amount each time, and time the pendulum for a minute at each length. Be sure to let the pendulum swing only a few inches from one extreme position to another. 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, and give the table in your response. Let the first column be the length, the second the frequency. Label this table Data Set 1.
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RESPONSE --> 1 FT = ABOUT 60 OSCILLATIONS PER MIN 2FT = ABOUT 30 OSCILLATIONS PER MIN 3 FT = ABOUT 15 OSCILLATIONS PER MIN 4 FT = ABOUT 7.5 OSCILLATIONS PER MIN
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17:47:54 Analyzing The Experiment Look at your data table for f vs. L 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?
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RESPONSE --> IT LOOKS LIKE THERE IS A CONSISTANT PATTERN OF INCREASING INCREASE.
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17:48:38 `q011. 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.
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RESPONSE --> OK.
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17:49:45 Describe the graph using the same type of language you used earlier. How does the graph reflect the behavior of the numbers on the table?
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RESPONSE --> IT GOES FROM LEFT TO RIGHT AT AN INCREASING RATE THAT IS CONSTANT. TAHT IS THE WAY THAT THE NUMBERS SEEM TO BE CALCULATING.
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17:50:11 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, a 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.
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RESPONSE --> OK
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17:50:38 `q012. Now sketch an accurate frequency vs. length graph of your results by marking a scale and plotting points, and compare this graph with your rough sketch. How does the shape of your graph compare with the shape of the rough sketch?
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RESPONSE --> ROUGH SKETCH IS JUST SLIGHTLY OFF. BUT THEY LOOK ABOUT THE SAME.
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17:52:17 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.
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RESPONSE --> ABOUT THESAME YET SLIGHTLY OFF.
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17:54:07 `q013. From your data, figure out the period associated with each length. Make a table of period vs. length, and label it clearly.
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RESPONSE --> EVERYTIME THE LEGNTH INCREASES THE OSCILLATIONS DOUBLE THEIR INCREASE.
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17:55:20 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 give the corrected table in your response. 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 from the graph you sketched before you made the table.
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RESPONSE --> THERE ISN'T THAT MUCH DIFFERENCE.
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17:55:39 `q014. Sketch your graph of period vs. length. Describe the graph (i.e., is it increasing or decreasing; and at an increasing or a decreasing rate?) .
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RESPONSE --> IT IS INCREASING AT A DECREASING RATE.
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17:55:54 The graph of period vs. length is increasing--the longer the pendulum the longer it takes to complete a cycle. The graph is increasing at a decreasing rate, since with every foot of increased length the period increases by a smaller and smaller amount. If your graph didn't show these characteristic then you should go back conceived where your observations and/or calculations when wrong and report any revised results here.
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RESPONSE --> OK
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17:56:12 06-19-2005 17:56:12 `q015. The formula for the period of the pendulum is very close to T = .2 `sqrt(L), with T in seconds when L is in centimeters. The exact formula is T = 2 `pi `sqrt( L / g), where L is the length of the pendulum and g the acceleration of gravity; however this formula will be developed later in the course and for now we will use T = .2 `sqrt(L), remembering that L must be measured in cm. What should be the period of the pendulum, according to this formula, for lengths of 1, 2, 3 and 4 feet (remember that to use T = .2 `sqrt(L) the length L must be in cm and that the period T will be in seconds)?
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NOTES -------> The formula for the period of the pendulum is very close to T = .2 `sqrt(L), with T in seconds when L is in centimeters. The exact formula is T = 2 `pi `sqrt( L / g), where L is the length of the pendulum and g the acceleration of gravity; however this formula will be developed later in the course and for now we will use T = .2 `sqrt(L), remembering that L must be measured in cm. What should be the period of the pendulum, according to this formula, for lengths of 1, 2, 3 and 4 feet (remember that to use T = .2 `sqrt(L) the length L must be in cm and that the period T will be in seconds)?
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17:56:20 06-19-2005 17:56:20 The periods will be 1.10 sec, 1.56 sec, 1.91 sec and 2.21 sec. If you didn't obtain these results then go back and substitute again in see if you do get the correct results. If not explain in detail how you are using the formula T = .2 `sqrt(L).
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NOTES ------->
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ÉçÃ{íÉìÅ…š½öÎោ£àÜ÷CÞhéÏ Student Name: assignment #001 001. Areas......!!!!!!!!...................................
21:48:29 `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 --> A = LW A = 3m*4m A = 12 cubic square m
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21:48:44 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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RESPONSE --> yes
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22:16:14 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE -->
To find the area: A = 1/2 BH A = 1/2 (4.0 * 3.0) A = 6 METERS SQUARED.................................................
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22:16:25 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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RESPONSE --> YES
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22:20:13 `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 --> A = BH 5.0 * 2.0 = 10.0 M2
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22:20:24 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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RESPONSE --> YES
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22:21:26 `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 BH A = 5.0 * 2.0 A = 10.0 SQUARE CM
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22:21:32 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 --> YES
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22:26:42 `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 --> A = 1/2 (B1 + B2) H A = 1/2 (4.0 + B2) 5.0 A =( 2 + 1/2B2 ) 5.0 A = 10.0 + 10.0B2
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22:27:33 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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RESPONSE --> OK SO I NEEDED TO DO A RETANGLE FORMULA.
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22:28:59 `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 --> A = LWH A = 4*3.0* 8.0 A = 96
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22:29:43 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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RESPONSE --> I DO NOT UNDERSTAND HOW THIS FORMULA CAME ABOUT BUT OK
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22:33:19 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> A = 3.14 * R2 A = 3.14 * 3.00 SQUARED A = 3.14 * 9 A = 28.6 CM2
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22:33:51 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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RESPONSE --> YEAH
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22:35:40 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C = 2 PI (R) C = 2 *3.14 * 3 C = 18.84 CM SQUARED
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22:35:58 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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RESPONSE --> YES I DID
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22:38:02 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> A = pi (R2) A = 3.14 (6 SQUARED) A = 3.14 ( 36) A = 113.04 M2
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22:38:50 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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RESPONSE --> OK I NEED TO APPROXIMATE MORE CAREFULLY.
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22:41:02 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> A = 14 * PI A = 13. 9822 A = 13.98 M2
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22:42:00 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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RESPONSE --> OK FIND THE AREA BY SOLVING TO FIND THE RADIUS FIRST
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22:45:11 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> A = PI (R2) 78 = PI (R2) 78/PI = PI/PI (R2) `SQRT 24.83 = `SQRT R2 4.98 M2 = RADIUS
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22:45:44 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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RESPONSE --> OK SO I ROUND OFF TO 5.
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22:46:15 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> WELL WE JUST SEE A BOX THAT IS WIDER THAN IT IS TALL.
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22:46:43 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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RESPONSE --> YES HYPOTHETICALLY SPEAKING
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22:48:29 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> THE AREA OF A RIGHT TRIANGLE IS A = 1/2 BH THAT IS THE BOTTOM OF THE TRIANGLE TIMES THE HEIGHT OF THE RIGHT ANGLE. HALF OF ALL OF THAT GIVES YOU THE SQUARE UNITS OF AREA.
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22:48:45 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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RESPONSE --> YES
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22:49:07 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> WE DO THE AREA OF A RETANGLE ...
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22:49:13 06-22-2005 22:49:13 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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22:49:19 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 --> OK
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22:49:26 06-22-2005 22:49:26 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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NOTES ------->
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22:49:37 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> WE DO IT LIKE A RETANGLE
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22:49:45 06-22-2005 22:49:45 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 ------->
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22:49:52 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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RESPONSE --> YEAH
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22:50:31 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> WE SAY THAT AREA IS EQUAL TO PIE BEING MULTIPLIED BY THE RADIUS SQUARED.
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22:50:37 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> YES
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22:52:25 `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 --> C = 2 PI * R JUST REMEMBER THAT THE CIRCUMFERENCE IS THE MEASUREMENT OF THE CIRCLE ALL THE WAY AROUND. THE SYMBOL FOR THIS IS PI BUT WE HAVE TO ACCOUNT FOR THE LENGTH OF THE DIAMETER.
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22:52:34 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 --> YES
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22:53:02 `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 APPLIED WHAT I KNOW AND WHAT I HAVE LEARNED TO MEMORY AND NOTES SAVED.
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