course Phy 201 sorry all this took so long-- I have had a few weeks out of town..... I will be sending several more assignments in the next couple of days (I am finishing up a couple simultaneously-- they sort of go with each other)also-- the text layout is somewhat confusing-- it's explanation of significant figures and the wording of some questions were not very helpful! Physics
......!!!!!!!!...................................
09:57:48 Briefly state what you think velocity is and how you think it is an example of a rate.
......!!!!!!!!...................................
RESPONSE --> velocity and rates are both ""change in some parameter"" in some unit of time..... In the case of velocity we are looking at change in the position of an object in some direction over a period of time. For instance a car changes its position 50 miles to the north in 1 hour.
.................................................
......!!!!!!!!...................................
09:58:04 ** A rate is a change in something divided by a change in something else. This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
10:01:51 Given average speed and time interval how do you find distance moved?
......!!!!!!!!...................................
RESPONSE --> If i have an average speed (50 miles an hour) and a time, (2 hours) then I would multipy the speed by the time interval. in this instance, 100 miles.
.................................................
......!!!!!!!!...................................
10:02:01 ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
10:04:04 Given average speed and distance moved how do you find the corresponding time interval?
......!!!!!!!!...................................
RESPONSE --> since speed * time= distance, then distance / speed= time. 25 miles/50 miles/hr= 1/2 hour.
.................................................
......!!!!!!!!...................................
10:05:28 ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
10:07:20 Given time interval and distance moved how do you get average speed?
......!!!!!!!!...................................
RESPONSE --> again you just rearrange. if distance/ speed is time, then distance/time is speed. so 100 miles in 4 hours= 25 miles/hour
.................................................
......!!!!!!!!...................................
10:07:29 ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................????m????f???
Student Name: assignment #002 002. Describing Graphs U???X??????x????? assignment #001 002. Describing Graphs ?w???????~???Student Name: assignment #001 001. Rates
......!!!!!!!!...................................
02:48:17 `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 --> to receive the next question in a QA, click on the next question button on the left hand side of the qa. You must write a reply in the right hand space and click the ""enter response button"" before you can click for the next question or answer. If you are replying to a question, write your best answer and reasoning in the right hand box. If you are replying to an answer a self critique or acknowledgement is appropriate. do not write in the left hand box. you may save your answers, self critiques or insights into a notes file in vhmthphy by clicking save as notes on the right hand side. Each time you are ready for the next question or answer, click next question/answer button on the left hand side.
.................................................
......!!!!!!!!...................................
02:48:27 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
02:49:14 `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 --> If you make 50 dollars in 5 hours, you are making 10 dollars per hour.
.................................................
......!!!!!!!!...................................
02:49:32 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
.................................................
......!!!!!!!!...................................
02:59:25 `q003.If you make $60,000 per year then how much do you make per month?
......!!!!!!!!...................................
RESPONSE --> 60,000 per year divided into 12 months is 5000 each month.
.................................................
......!!!!!!!!...................................
02:59:37 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
.................................................
......!!!!!!!!...................................
03:08:17 `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?
......!!!!!!!!...................................
RESPONSE --> the business makes an average of 5000 per month. It is very unlikely that a business (not a salaried individual) would make exactly the same amount each month. Therefore, one would expect larger or smaller amounts on any given month.
.................................................
......!!!!!!!!...................................
03:08:35 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 --> exactly
.................................................
......!!!!!!!!...................................
03:12:29 `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?
......!!!!!!!!...................................
RESPONSE --> you are averaging 50 miles an hour (300miles/6hrs= 50miles/1hr) again, it is an average because at any given point in time you are going slower or faster due to traffic conditions, weather, etc.
.................................................
......!!!!!!!!...................................
03:12:56 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
.................................................
......!!!!!!!!...................................
03:14:12 `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?
......!!!!!!!!...................................
RESPONSE --> 1200 miles/60gallons= 20miles/1gallon or 20 miles per gallon
.................................................
......!!!!!!!!...................................
03:20:51 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 --> ok, I did miles per gallon. I did not understand exactly what you were asking for.... I will remember to check the order of terms in future questions.
.................................................
......!!!!!!!!...................................
03:29:10 `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?
......!!!!!!!!...................................
RESPONSE --> we have the final quantity (300 mile, 60000 dollars) and we have the number (n) but we do not have the individual values. We can calculate the average if we have the sum and the dividend (hours, months, etc) that we choose to use.
.................................................
......!!!!!!!!...................................
03:29:33 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
.................................................
......!!!!!!!!...................................
04:27:15 `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?
......!!!!!!!!...................................
RESPONSE --> first group averaged a strength of 14.7lbs/pushup the second group averaged 162/50 or 3.2 lbs/pushup. since we have no way of knowing what the original lifting strength was, I am using the first group as a control group. with an increase of 40 pushups per day and an increase of 15 lbs of lifting weight over the 1st group, we can say that the average lifting strength increased at 15lbs/40 pushups or .375 lbs/pushup
.................................................
......!!!!!!!!...................................
04:27:59 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 --> exactly.
.................................................
......!!!!!!!!...................................
04:32:34 `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?
......!!!!!!!!...................................
RESPONSE --> here, the number of pushups is irrelevent since it cancels out, instead, we use the added weight. the second group gained 17 lbs of lifting weight for 20lbs of shoulder weight. therefore 17lbslw/20lbssw or .85lbs of lifting weight per lb of shoulder weight
.................................................
......!!!!!!!!...................................
04:32:48 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 --> exactly
.................................................
......!!!!!!!!...................................
04:53:55 `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?
......!!!!!!!!...................................
RESPONSE --> the runner takes 10 seconds to cover the 100 meters between the first and second positions. therefore he is traveling at an average rate of 100m/10sec or 10 m/sec between the 100 and 200 meter marks.
.................................................
......!!!!!!!!...................................
04:54:08 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
.................................................
......!!!!!!!!...................................
04:59:10 `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?
......!!!!!!!!...................................
RESPONSE --> if we average the two speeds, we get an average rate of 9.5 meters per second for 100 meters. by dividing, we see that this is about 10.5 seconds to cover the distance between the two marks.
.................................................
......!!!!!!!!...................................
05:15:33 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 --> ok
.................................................
......!!!!!!!!...................................
05:32:39 `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?
......!!!!!!!!...................................
RESPONSE --> in this instance, we have two individual velocities, but not the speed between the two marks. to find the best approximate speed, it makes sense to average the two speeeds I have over the distance.
.................................................
......!!!!!!!!...................................
05:34:08 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, I see the fact that the earlier quantities were not rates. I also see that I needed one rate, not two to calculate the answer to time passed over a distance.
.................................................
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.
......!!!!!!!!...................................
RESPONSE --> h*w= 3*4=12m^2
.................................................
......!!!!!!!!...................................
03:22:09 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
03:23:37 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
......!!!!!!!!...................................
RESPONSE --> 1/2 base*height or 2*3=6 m^2 (meters *meters = square meters...)
.................................................
......!!!!!!!!...................................
03:23:51 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
03:29:19 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
......!!!!!!!!...................................
RESPONSE --> no matter what the original shape of a given parallelogram, it is possible to ""rearrange"" any offset triangles to form a square or rectangle. the base and height are then multiplied as in a standard rectangle. 5*2= 10m^2
.................................................
......!!!!!!!!...................................
03:29:34 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.
......!!!!!!!!...................................
RESPONSE --> yup
.................................................
......!!!!!!!!...................................
03:30:18 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
......!!!!!!!!...................................
RESPONSE --> 1/2 base * h= 5m^2 (1/2*2*5)
.................................................
......!!!!!!!!...................................
03:31:25 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.
......!!!!!!!!...................................
RESPONSE --> yes-- these formulae are so ingrained that I forget to justify why i use them at this point........
.................................................
......!!!!!!!!...................................
05:55:58 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
......!!!!!!!!...................................
RESPONSE --> a trapezoid can be divided into two triangles. since each triangle has the same ""height"" then we can say that the trapezoid is the sum of (1/2)(b1)h + (1/2)(b2)h which can be changed to 1/2(b1+b2)h. since 1/2 of b1 + b2 is the same as the AVERAGE of the two bases, then the area of this trapezoid is 5*4 or 20 km^2.
.................................................
......!!!!!!!!...................................
05:56:23 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.
......!!!!!!!!...................................
RESPONSE --> okay, i used a different method.....
.................................................
......!!!!!!!!...................................
06:04:48 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
......!!!!!!!!...................................
RESPONSE --> 1/2 * (3+8)*4 (using the addition of triangles again) = 22cm^2. You can prove this by taking the area of each triangle (1/2*3*4) and (1/2*8*4) and adding them. so... 6+ 16= 22 cm ^2.....
.................................................
......!!!!!!!!...................................
06:05:10 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.
......!!!!!!!!...................................
RESPONSE --> again, a different method with the same result.....
.................................................
......!!!!!!!!...................................
23:34:33 `q007. What is the area of a circle whose radius is 3.00 cm?
......!!!!!!!!...................................
RESPONSE --> ok, i couldn't exactly remember the formula... ( i always mix up circumference and area...) so, think of a circle in a square. the complete width of the square would be 6X6 ( the diameter of the circle), so the area of the square would be 36 cm ^2. The circle takes up less space, but he width is the same 6. we use pi in circle formulae. I remember 2 pi r2 and pi r^2. the first is impossible because the area of the circle would be larger than the area of the square enclosing it. pi * 3^2 is 28.274 which is a reasonable answer
.................................................
......!!!!!!!!...................................
23:35:46 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.
......!!!!!!!!...................................
RESPONSE --> ok... exept for the number of significant figures....
.................................................
......!!!!!!!!...................................
23:38:04 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
......!!!!!!!!...................................
RESPONSE --> circumference is 2 pi r or 6pi cm. in decimals this would be ~ 3.14 * 6 or 18.84 cm
.................................................
......!!!!!!!!...................................
23:38:21 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:40:22 `q009. What is the area of a circle whose diameter is exactly 12 meters?
......!!!!!!!!...................................
RESPONSE --> the radius is 6 so 6^2 * pi or 12 pi cm^2. or ~ 37.7
.................................................
......!!!!!!!!...................................
23:41:53 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.
......!!!!!!!!...................................
RESPONSE --> 6 ^ 2 is 36 not 12--- duhhhhhh
.................................................
......!!!!!!!!...................................
22:46:32 `q010. What is the area of a circle whose circumference is 14 `pi meters?
......!!!!!!!!...................................
RESPONSE --> work backwards. 2pir is circumference. divide 14pi by 2pi and you have a radius of 7. pi r ^2= pi * (7^2)= 49pi cm ^2
.................................................
......!!!!!!!!...................................
22:46:54 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:56:03 `q011. What is the radius of circle whose area is 78 square meters?
......!!!!!!!!...................................
RESPONSE --> area is pi r^2 so, again, we work bacwards. r^2* pi = 78. 78/pi=r^2 sqrt(78/pi)= r r=4.98 to check, (4.98)^2* pi= 78
.................................................
......!!!!!!!!...................................
22:56:35 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:11:28 `q012. Summary Question 1: How do we visualize the area of a rectangle?
......!!!!!!!!...................................
RESPONSE --> i always tell my students to think of a tile floor in a room. in a rectangular room, there will be x number of tiles running the length of the room and y number of tiles running the width (or x rows of y tiles). so if we multiply, we have (for example) 7 rows of 8 tiles apiece or 54 tiles altogether. the room is then 54 square tiles. (whether the tiles are 1 ft square, 1 meter square or whatever)
.................................................
......!!!!!!!!...................................
23:11:41 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.
......!!!!!!!!...................................
RESPONSE --> exactly
.................................................
......!!!!!!!!...................................
23:13:44 `q013. Summary Question 2: How do we visualize the area of a right triangle?
......!!!!!!!!...................................
RESPONSE --> a right triangle is basically 1/2 of a rectangle. since a rectangle's area is l*w. then the triangles area is 1/2 of l*w or 1/2l*w
.................................................
......!!!!!!!!...................................
23:13:57 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.
......!!!!!!!!...................................
RESPONSE --> exactly
.................................................
......!!!!!!!!...................................
23:20:26 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
......!!!!!!!!...................................
RESPONSE --> no matter what the original shape of a given parallelogram, it is possible to ""rearrange"" any offset triangles to form a square or rectangle. the base and height are then multiplied as in a standard rectangle.
.................................................
......!!!!!!!!...................................
23:20:36 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:21:57 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
......!!!!!!!!...................................
RESPONSE --> 1/2 (b1+b2) * h
.................................................
......!!!!!!!!...................................
23:22:06 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:22:40 `q016. Summary Question 5: How do we calculate the area of a circle?
......!!!!!!!!...................................
RESPONSE --> the area of a circle is the square of the radius multiplied by pi
.................................................
......!!!!!!!!...................................
23:22:48 We use the formula A = pi r^2, where r is the radius of the circle.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
23:23:55 `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?
......!!!!!!!!...................................
RESPONSE --> the circumference is 2*pi*radius. since it has no squared units it is linear (the ""length"" around the circle)
.................................................
......!!!!!!!!...................................
23:24:02 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.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
23:29:29 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> This was a review of some of the basic math I will be using in this course. Instead of JUST using formula, vizualizing why the formulae are used helps to see when different formula will be useful. as reminders, I have written the formulae down in a ""glossary"" for reference.
.................................................
......!!!!!!!!...................................
23:29:34 This ends the first assignment.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:29:39 002. Volumes
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:31:53 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
......!!!!!!!!...................................
RESPONSE --> or, 5 rows of 7cm stacked 3 high. this gives me 5*7 or 35 square cm stacked 3cm high------ 105 cm^3
.................................................
......!!!!!!!!...................................
23:32:32 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
......!!!!!!!!...................................
RESPONSE --> you switched between cm and m, other than that, ok
.................................................
......!!!!!!!!...................................
23:33:14 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
......!!!!!!!!...................................
RESPONSE --> 48 m^2* 2 m= 96 m^3
.................................................
......!!!!!!!!...................................
23:33:26 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
......!!!!!!!!...................................
RESPONSE --> exactly
.................................................
......!!!!!!!!...................................
23:36:07 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
......!!!!!!!!...................................
RESPONSE --> using the same principle as the volume of a cube, we take the ""area"" of the base and then ""stack"" them. so 20 m^2 stacked 40 m high or 20 m^2 * 40 m= 800 m^3
.................................................
......!!!!!!!!...................................
23:36:19 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:38:12 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
......!!!!!!!!...................................
RESPONSE --> first we find the area of the base. pi * (5cm)^2= 25 pi cm^2 then we multiply by the height-- 30* 25* pi= 750pi cm^3
.................................................
......!!!!!!!!...................................
23:38:13 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
23:38:33 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:44:20 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
......!!!!!!!!...................................
RESPONSE --> ok so tomato can that is ~ 10 cm diameter and 15 cm tall would be 25cm^2*15cm= 375 cm^3
.................................................
......!!!!!!!!...................................
23:44:56 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:48:31 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
......!!!!!!!!...................................
RESPONSE --> i had forgotten the formula for pyramids, but in all other volume operations, we have taken the area of one side and multiplied it by an extra dimension. the area of the base is 50 times 1/2 the altitude of 60 would be 1500 cm^3
.................................................
......!!!!!!!!...................................
21:49:11 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
......!!!!!!!!...................................
RESPONSE --> 1/3----- of course!
.................................................
......!!!!!!!!...................................
07:08:03 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
......!!!!!!!!...................................
RESPONSE --> a cone is 1/3 of a cylinder ( i thought 1/2, but I realized that would actually give me a rectangle as one face....) 1/3(9)(20)= 60 m^3 (1/3 *height*Area)
.................................................
......!!!!!!!!...................................
07:08:18 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
......!!!!!!!!...................................
RESPONSE --> yeah!
.................................................
......!!!!!!!!...................................
22:11:31 `q008. What is a volume of a sphere whose radius is 4 meters?
......!!!!!!!!...................................
RESPONSE --> V= 4/3pi(r^3) so 64*pi*4/3= 85 1/3 pi m^3
.................................................
......!!!!!!!!...................................
22:12:00 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
......!!!!!!!!...................................
RESPONSE --> divided out, 85 1/3 pi...
.................................................
......!!!!!!!!...................................
22:21:20 `q009. What is the volume of a planet whose diameter is 14,000 km?
......!!!!!!!!...................................
RESPONSE --> this is a sphere problem. 4/3 pi (7000^3)=4/3 pi (343000000000)=457333333333 1/3 pi km^3
.................................................
......!!!!!!!!...................................
22:21:47 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:07:19 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
......!!!!!!!!...................................
RESPONSE --> we think of ""stacking"" circles of the same size, multiplying the height of the cylinder by the area of the circle.
.................................................
......!!!!!!!!...................................
21:07:34 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:10:48 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
......!!!!!!!!...................................
RESPONSE --> the idea is that a cone or pyramid is 1/3 of the structure it is carved out of (cylinder or rectangular prism) so, you would multiply 1/3 A*h where A is the area of the base and h is the altitude of the object.
.................................................
......!!!!!!!!...................................
21:10:59 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
21:11:13 `q012. Summary Question 3: What is the formula for the volume of a sphere?
......!!!!!!!!...................................
RESPONSE --> 4/3 pi r^3
.................................................
......!!!!!!!!...................................
21:11:19 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
21:12:27 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> In several cases i used visuals to help me understand the principles. i have typed the formulas and some notes (why this is the formula....) into a glossary I keep on the computer.
.................................................
......!!!!!!!!...................................
21:12:36 This ends the second assignment.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:12:44 003. Misc: Surface Area, Pythagorean Theorem, Density
......!!!!!!!!...................................
RESPONSE --> start
.................................................
......!!!!!!!!...................................
21:17:12 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
......!!!!!!!!...................................
RESPONSE --> surface area is simply adding the areas of the different faces of a solid object. so in this case a rectangular solid has six faces (top. bottom. and 4 sides.) assuming the altitude is the 6 meter measurement, we have two faces of 12 m^2 , 2 faces of 3x6 or 18m ^2 and two faces of 4x6 or 24 m ^2. this gives us 24 + 36 + 48= 108 m^2
.................................................
......!!!!!!!!...................................
21:17:32 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
......!!!!!!!!...................................
RESPONSE --> exactly
.................................................
......!!!!!!!!...................................
22:43:38 `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
......!!!!!!!!...................................
RESPONSE --> the surface area of the sides is 2*pi*r*l= 5*2*12*pi= 120pi m^2 if the cylinder is closed, then we would add the surface area of the two bases-- 5^2*pi*2= 50 pi + 120 pi= 170pi m^2
.................................................
......!!!!!!!!...................................
22:50:05 The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.
......!!!!!!!!...................................
RESPONSE --> This is an easier way to think about it than just using the formula....
.................................................
......!!!!!!!!...................................
22:54:48 `q003. What is surface area of a sphere of diameter three cm?
......!!!!!!!!...................................
RESPONSE --> 4pi*r^2= the surface area of a sphere..... the radius is 1/2 of the diameter. 1.5^2= 2.25 * 4=9*pi= 9pi
.................................................
......!!!!!!!!...................................
22:55:00 The surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
22:58:46 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
......!!!!!!!!...................................
RESPONSE --> a^2 + b^2 = c^2 (the pythagorean theorem is tatooed on my brain!) 25 + 81= 106 m^2= 10.296
.................................................
......!!!!!!!!...................................
22:59:03 The Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.
......!!!!!!!!...................................
RESPONSE --> yes...
.................................................
......!!!!!!!!...................................
23:01:12 `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?
......!!!!!!!!...................................
RESPONSE --> 36= 16m + x^2 x^2=20 x~4.5
.................................................
......!!!!!!!!...................................
23:01:28 If c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
23:23:10 `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?
......!!!!!!!!...................................
RESPONSE --> density is a function of mass and volume so, 700g/12cm= 58 1/3 g/cm^3
.................................................
......!!!!!!!!...................................
23:27:21 The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).
......!!!!!!!!...................................
RESPONSE --> ok--- it would help if I calculated the volume instead of just using one dimension.
.................................................
......!!!!!!!!...................................
23:37:16 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
......!!!!!!!!...................................
RESPONSE --> again, density is a function of mass/volume. since the density is 3000 kg/m^3 and the volume is 4/3 pi r^3, then the 3000 kg/m^3= m/(4/3 pi (4^3)) 3000 kg/m^3* (4/3 * 64* pi m^3) 3000 kg/m^3 (85 1/3 pi m^3)= 256000 kg
.................................................
......!!!!!!!!...................................
23:39:34 A average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures.
......!!!!!!!!...................................
RESPONSE --> I left out pi..... I am being very inattentive on this assignment.
.................................................
......!!!!!!!!...................................
23:50:01 `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?
......!!!!!!!!...................................
RESPONSE --> if we think of density as a proportion, then D= m/v where d is density, m is mass and v is volume, then we can say that the mass of the first object is 24 g and the mass of the second object is v*d or 20 grams. the created object has a volume of 16 cm^3 and a mass of 44 grams. 44 grams/16 cm^3= 2.75 g/cm^3
.................................................
......!!!!!!!!...................................
23:50:11 The first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
00:01:14 `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?
......!!!!!!!!...................................
RESPONSE --> again density = m/v the dimensions of the box are irrelevent since you have the given the volume of both items inside. 2100kg/m^3 *27m^3= 56700 kg of sand, + 8000 kg/m^3*3 m^3 or 24000 kg of cannonballs = 80700kg of material and a volume of 30 cubic meters. the density is 80700/30 or 2690 kg/m^3
.................................................
......!!!!!!!!...................................
00:01:36 We find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:11:29 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
......!!!!!!!!...................................
RESPONSE --> 1700000 m^2 * .015 m= 25500 m^3 in the slick. the mass of the slick is 860kg/m^3 * 25500 m^3= 21930000 kg.
.................................................
......!!!!!!!!...................................
00:13:45 The volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:22:32 `q011. Summary Question 1: How do we find the surface area of a cylinder?
......!!!!!!!!...................................
RESPONSE --> Surface area = circumference of base (2*pi*r)* height= 2pi*r*h + 2 pi r^2 or, the surface area of each of the two circles making up the bases added to the surface area of the sides.
.................................................
......!!!!!!!!...................................
00:32:26 The curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:34:38 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
......!!!!!!!!...................................
RESPONSE --> Surface area: A= 4* pi* r^2
.................................................
......!!!!!!!!...................................
00:34:46 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:50:53 `q013. Summary Question 3: What is the meaning of the term 'density'.
......!!!!!!!!...................................
RESPONSE --> density is a measurement of mass per volume
.................................................
......!!!!!!!!...................................
00:51:06 The average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
00:57:52 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
......!!!!!!!!...................................
RESPONSE --> since density is mass per unit of volume ( or m/v=d) then density * volume = mass, and mass/density= volume
.................................................
......!!!!!!!!...................................
01:00:03 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:00:59 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I saved notes and examples in a file, and I added the various formulae to a glossary....
.................................................
......!!!!!!!!...................................
01:01:06 This ends the third assignment.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:01:15 004. Units of volume measure
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:02:13 `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
......!!!!!!!!...................................
RESPONSE --> 10*10*10= 1000 cm^3
.................................................
......!!!!!!!!...................................
01:02:21 The volume of the container is 10 cm * 10 cm * 10 cm = 1000 cm^3. So it would take 1000 cubic centimeters of fluid to fill the container.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:14:49 `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
......!!!!!!!!...................................
RESPONSE --> since a meter is a 100 cm, then 10 cubes end to end would create a 10 cubes wide by 10 cubes long stacked 10 cubes high or 10* 10*10= 1000 cubes
.................................................
......!!!!!!!!...................................
01:15:03 It takes ten 10 cm cubes laid side by side to make a row 1 meter long or a tower 1 meter high. It should therefore be clear that the large cube could be built using 10 layers, each consisting of 10 rows of 10 small cubes. This would require 10 * 10 * 10 = 1000 of the smaller cubes.
......!!!!!!!!...................................
RESPONSE --> yep
.................................................
......!!!!!!!!...................................
23:35:14 `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
......!!!!!!!!...................................
RESPONSE --> a kilometre is 100 metres. so it would take 100 tiles in a row to take up a kilometre. so 100 of such rows would make up a square km. 100 * 100 = 10,000 tiles
.................................................
......!!!!!!!!...................................
23:35:51 It takes 1000 meters to make a kilometer (km). To cover a square 1 km on a side would take 1000 rows each with 1000 such tiles to cover 1 square km. It therefore would take 1000 * 1000 = 1,000,000 squares each 1 m on a side to cover a square one km on a side. We can also calculate this formally. Since 1 km = 1000 meters, a square km is (1 km)^2 = (1000 m)^2 = 1,000,000 m^2.
......!!!!!!!!...................................
RESPONSE --> again, i din't check definitions.....
.................................................
......!!!!!!!!...................................
23:38:12 `q004. How many cubic centimeters are there in a liter?
......!!!!!!!!...................................
RESPONSE --> there are 1000 cubic centimeters in a litre
.................................................
......!!!!!!!!...................................
23:38:18 A liter is the volume of a cube 10 cm on a side. Such a cube has volume 10 cm * 10 cm * 10 cm = 1000 cm^3. There are thus 1000 cubic centimeters in a liter.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:45:19 `q005. How many liters are there in a cubic meter?
......!!!!!!!!...................................
RESPONSE --> a litre is 1000 cm^3 a cubic metre is 100 cm long by 100 wide by 100 high, or 1,000,000 cm^3 1000000/1000= 1000 litres
.................................................
......!!!!!!!!...................................
23:45:34 A liter is the volume of a cube 10 cm on a side. It would take 10 layers each of 10 rows each of 10 such cubes to fill a cube 1 meter on a side. There are thus 10 * 10 * 10 = 1000 liters in a cubic meter.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
23:46:58 `q006. How many cm^3 are there in a cubic meter?
......!!!!!!!!...................................
RESPONSE --> as i said in my earlier answer it would be 1 million cm^3. 100 long x 100 high x 100 wide
.................................................
......!!!!!!!!...................................
23:47:08 There are 1000 cm^3 in a liter and 1000 liters in a m^3, so there are 1000 * 1000 = 1,000,000 cm^3 in a m^3. It's important to understand the 'chain' of units in the previous problem, from cm^3 to liters to m^3. However another way to get the desired result is also important: There are 100 cm in a meter, so 1 m^3 = (1 m)^3 = (100 cm)^3 = 1,000,000 cm^3.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:16:42 `q007. If a liter of water has a mass of 1 kg the what is the mass of a cubic meter of water?
......!!!!!!!!...................................
RESPONSE --> since there are 1000 liters in a cubic metre, then 1kg * 1000= 1000 kg
.................................................
......!!!!!!!!...................................
21:17:05 Since there are 1000 liters in a cubic meter, the mass of a cubic meter of water will be 1000 kg. This is a little over a ton.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:46:54 `q008. What is the mass of a cubic km of water?
......!!!!!!!!...................................
RESPONSE --> a cubic kilometer contains 1000 x1000 x 1000 cubic m since m^3 is 1000 kg, 1km^3= 1000^4 or 10^12 kg
.................................................
......!!!!!!!!...................................
21:47:13 A cubic meter of water has a mass of 1000 kg. A cubic km is (1000 m)^3 = 1,000,000,000 m^3, so a cubic km will have a mass of 1,000,000,000 m^3 * 1000 kg / m^3 = 1,000,000,000,000 kg. In scientific notation we would say that 1 m^3 has a mass of 10^3 kg, a cubic km is (10^3 m)^3 = 10^9 m^3, so a cubic km has mass (10^9 m^3) * 1000 kg / m^3 = 10^12 kg.
......!!!!!!!!...................................
RESPONSE --> cool
.................................................
......!!!!!!!!...................................
21:56:11 07-01-2006 21:56:11 `q009. If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water?
......!!!!!!!!...................................
NOTES -------> ok, a cubic kilometer contains 1000 x1000 x 1000 cubic m Since there are 1000 liters in a cubic meter, then there are 1000000000000 liters in a cubic km. 5000000000 people * 2 l a day= 10000000000 l daily 10^12/10^10= 10^2 or 100 days
.......................................................!!!!!!!!...................................
21:56:21 `q009. If each of 5 billion people drink two liters of water per day then how long would it take these people to drink a cubic km of water?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:57:18 5 billion people drinking 2 liters per day would consume 10 billion, or 10,000,000,000, or 10^10 liters per day. A cubic km is (10^3 m)^3 = 10^9 m^3 and each m^3 is 1000 liters, so a cubic km is 10^9 m^3 * 10^3 liters / m^3 = 10^12 liters, or 1,000,000,000,000 liters. At 10^10 liters per day the time required to consume a cubic km would be time to consume 1 km^3 = 10^12 liters / (10^10 liters / day) = 10^2 days, or 100 days. This calculation could also be written out: 1,000,000,000,000 liters / (10,000,000,000 liters / day) = 100 days.
......!!!!!!!!...................................
RESPONSE --> the program did something funny, but that is the answer i entered.
.................................................
......!!!!!!!!...................................
22:11:19 `q010. The radius of the Earth is approximately 6400 kilometers. What is the surface area of the Earth? If the surface of the Earth was covered to a depth of 2 km with water that what would be the approximate volume of all this water?
......!!!!!!!!...................................
RESPONSE --> Surface area: A= 4* pi* r^2 so A= 4pi* (6400^2) A = 4 * pi * 40960000 = 163,840,000pi km^2 = ~ 514718540.4 km ^2 Volume: V= 4/3*pi*r^3 actually this is two volumes.... the volume of the sphere 6402 radius - the volume 6400 radius would give us the volume, or the surface area * 2 km(height) would give us the volume required. since I already have the surface area, then 2 * 16840000 pi = 327680000 pi km^3 or ~ 1029437081 km ^3
.................................................
......!!!!!!!!...................................
22:12:58 The surface area would be A = 4 pi r^2 = 4 pi ( 6400 km)^2 = 510,000,000 km^2. A flat area of 510,000,000 km^2 covered to a depth of 2 km would indicate a volume of V = A * h = 510,000,000 km^2 * 2 km = 1,020,000,000 km^3. However the Earth's surface is curved, not flat. The outside of the 2 km covering of water would have a radius 2 km greater than that of the Earth, and therefore a greater surface area. However a difference of 2 km in 6400 km will change the area by only a fraction of one percent, so the rounded result 1,020,000,000,000 km^3 would still be accurate.
......!!!!!!!!...................................
RESPONSE --> so i would have been more accurate calculating both volumes and subtracting.....
.................................................
......!!!!!!!!...................................
22:16:53 `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
......!!!!!!!!...................................
RESPONSE --> A liter is the volume of a cube 10 cm on a side. So we can think of a litre as a cube 10 cm * 10 cm * 10 cm = 1000 cm^3. 1000 cubic centimeters in a liter.
.................................................
......!!!!!!!!...................................
22:17:01 Since a liter is a cube 10 cm on a side, we visualize 10 layers each of 10 rows each of 10 one-centimeter cubes, for a total of 1000 1-cm cubes. There are 1000 cubic cm in a liter.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:19:31 `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
......!!!!!!!!...................................
RESPONSE --> since a litre is a cube that is 10 cm on a side, and there are 100 cm in a meter. then 10 litres in a row, 10 rows stacked ten high gives me a cube 10* 10 * 10 or 1000 litres in a cubic meter
.................................................
......!!!!!!!!...................................
22:19:39 Since a liter is a cube 10 cm on a side, we need 10 such cubes to span 1 meter. So we visualize 10 layers each of 10 rows each of 10 ten-centimeter cubes, for a total of 1000 10-cm cubes. Again each 10-cm cube is a liter, so there are 1000 liters in a cubic meter.
......!!!!!!!!...................................
RESPONSE --> yes
.................................................
......!!!!!!!!...................................
22:21:20 `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
......!!!!!!!!...................................
RESPONSE --> since there are 1000 cm^3 in a cubic litre and 1000 l^3 in a m^3 then 1000 * 1000= 10 ^ 6 cm^3 in a m^3
.................................................
......!!!!!!!!...................................
22:21:58 One way is to know that there are 1000 liters in a cubic meters, and 1000 cubic centimeters in a cubic meter, giving us 1000 * 1000 = 1,000,000 cubic centimeters in a cubic meter. Another is to know that it takes 100 cm to make a meter, so that a cubic meter is (100 cm)^3 = 1,000,000 cm^3.
......!!!!!!!!...................................
RESPONSE --> yes, I have explained it both ways in this qa
.................................................
......!!!!!!!!...................................
22:30:32 `q014. Summary Question 4: There are 1000 meters in a kilometer. So why aren't there 1000 cubic meters in a cubic kilometer? Or are there?
......!!!!!!!!...................................
RESPONSE --> think of stacking blocks. 1000 cubes of a meter is a kilometer long, but only 1 m wide so you make 1000 rows to get it 1 km wide as well. It is still only 1 m high, so you have to stack your cubes 1000 high, now, you have a cube made up of smaller cubes, 1000 high by 1000 long by 1000 wide= 1000* 1000 * 1000= 1000000000 cubes
.................................................
......!!!!!!!!...................................
22:30:47 A cubic kilometer is a cube 1000 meters on a side, which would require 1000 layers each of 1000 rows each of 1000 one-meter cubes to fill. So there are 1000 * 1000 * 1000 = 1,000,000,000 cubic meters in a cubic kilometer. Alternatively, (1 km)^3 = (10^3 m)^3 = 10^9 m^3, not 1000 m^3.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:31:48 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> again, I added notes to a file on the computer and added formula to a glossary I am compiling.
.................................................
......!!!!!!!!...................................
22:31:55 This ends the fourth assignment.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"