course Phy201 |՝PhݰqjϠñassignment #001
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15:23:11 `questionNumber 10001 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> I think so
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15:28:10 `questionNumber 10001 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> Found the send file
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15:32:46 `questionNumber 10002 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> Each time I enter an answer by clicking on the Enter Response button, an answer that is typed in the ""answer box"" is submitted into the send file. I will never erase anything in the send file. Once it's there, it stays there, because the program never unwrites anything- even in the event of a ""crash"".
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15:36:18 `questionNumber 10002 Any time the instructor does not post a response to your access site by the end of the following day, you should resubmit your work using the Submit Work form, and be sure at the beginning to indicate that you are resubmitting, and also indicate the date on which you originally submitted your work. If you don't know where your access site is or how to access it, go to http://www.vhcc.edu/dsmith/_vti_bin/shtml.dll/request_access_code.htm and request one now. You can submit the q_a_prelim without your access code, but other assignments should contain your code.
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RESPONSE --> I requested a code on Friday, but haven't received one yet. I'll try to submit again.
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15:43:12 `questionNumber 10003 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> If I use a VHCC computer for my work submissions, I need to be extra careful to save my work, (ie. junk drive) or either by emailing it to myself. It's a good idea to save all work anyway.
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xj]yxܢ assignment #001 001. Areas qa areas volumes misc 09-14-2008
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21:43:13 `questionNumber 10000 `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=1/2 (b)(h) 1/2(4m)(3m) 1/2 x 12m2 6m2 confidence assessment: 2
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21:56:04 `questionNumber 10000 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 --> I answered the previous question as 6m^2, but hit response instead of clear box. Sorry! Didn't know how to go back. Ok. I understand so far. self critique assessment: 2
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21:57:45 `questionNumber 10000 `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 --> 5mx2m= 10m^2 confidence assessment: 3
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21:58:37 `questionNumber 10000 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 --> ok self critique assessment: 3
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22:01:07 `questionNumber 10000 `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 --> 1/2(bxh)=area 1/2(5.0cmx2.0cm)= 5cm^2 confidence assessment: 3
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22:02:09 09-14-2008 22:02:09 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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NOTES ------->
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22:05:43 `questionNumber 10000 `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 --> area=altitudexwidth 5kmx4km= 20km^2 confidence assessment: 2
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22:06:24 `questionNumber 10000 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 self critique assessment: 2
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22:13:45 `questionNumber 10000 `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 --> area=displacement=(Vf+Vo/2)(change t) = (8cm+3cm/2)(4cm) =(5.5cm)(4cm) =22cm^2 confidence assessment: 2
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22:14:26 `questionNumber 10000 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 --> ok I feel better now. self critique assessment:
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22:20:14 `questionNumber 10000 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> area= 3.14x r^2 = 3.14x 3.00cm^2 = 28.26cm^2 confidence assessment: 3
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22:21:58 `questionNumber 10000 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 --> Thanks for pointing that out. I'll try to be careful. self critique assessment: 2
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22:24:34 `questionNumber 10000 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> circumference= 2(3.14x 3cm) = 18.84cm confidence assessment: 2
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22:25:12 `questionNumber 10000 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 --> ok That makes sense self critique assessment: 3
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22:28:08 `questionNumber 10000 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> area= (3.14x d^2)/4 =(3.14x 144m^2)/4 =113.04m^2 confidence assessment: 3
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22:30:39 `questionNumber 10000 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. self critique assessment: 3
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22:35:36 `questionNumber 10000 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> If circumference = 14 pi meters, then the diameter must =14m, so radius=7m area= pi r^2 = (3.14)(49m^2) =153.86m^2 confidence assessment: 1
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22:36:41 `questionNumber 10000 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 self critique assessment: 2
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22:42:17 `questionNumber 10000 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> 78m^2= pi(r2) 78m^2x pi=r^2 244.92=r^2 15.65=r confidence assessment: 1
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22:44:05 `questionNumber 10000 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 --> I see. I multiplied both by pi, instead of divide. Yikes! self critique assessment:
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22:44:42 `questionNumber 10000 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> length x width confidence assessment: 3
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22:45:07 `questionNumber 10000 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 --> ok self critique assessment: 3
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22:45:48 `questionNumber 10000 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> 1/2 bxh confidence assessment: 3
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22:46:42 `questionNumber 10000 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> area= b x h confidence assessment: 3
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22:47:01 `questionNumber 10000 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 self critique assessment: 3
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22:48:30 `questionNumber 10000 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> area= displacement= avg. of the heightx width confidence assessment: 3
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22:48:48 `questionNumber 10000 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 --> ok self critique assessment: 3
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22:49:15 `questionNumber 10000 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> area= pix r^2 confidence assessment: 3
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22:49:26 `questionNumber 10000 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok self critique assessment: 3
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22:50:30 `questionNumber 10000 `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= pi x diameter area= a squared answer, circumference isn't squared confidence assessment: 3
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22:50:46 `questionNumber 10000 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 --> ok self critique assessment: 3
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22:52:43 `questionNumber 10000 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> studied my ass off. I have very little knowledge to organize, so it's simple. I've reviewed my notes after each class and done all of the assignments. That's all I know to do. confidence assessment: 2
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Ovلy튬rʑ씪 assignment #002 002. Volumes qa areas volumes misc 09-14-2008
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23:01:48 `questionNumber 20000 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> v= l x w x h = 3cmx 5cm x 7cm =105 cm^3 confidence assessment: 3
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23:05:51 09-14-2008 23:05:51 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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NOTES -------> Volume of any rectangular solid = Area x height ok.
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23:07:47 `questionNumber 20000 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> volume= area x ht = 48 m^2 x 2m = 96 m^3 = volume confidence assessment: 2
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23:08:44 `questionNumber 20000 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> ok. I hit 98 instead of 96 on the answer. Read my calculator wrong. self critique assessment: 2
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23:11:28 `questionNumber 20000 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> I have no idea? The answer will be m^3 confidence assessment: 0
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23:12:44 09-14-2008 23:12:44 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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NOTES -------> V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.
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23:14:59 `questionNumber 20000 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> v=pi r^2 x length =3.14x 5cm^2 x 30cm = 471 cm^3 confidence assessment: 2
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23:17:03 `questionNumber 20000 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies. The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.
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RESPONSE --> self critique assessment: 0
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23:17:41 `questionNumber 20000 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> volume= area x height confidence assessment: 2
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23:19:06 `questionNumber 20000 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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RESPONSE --> ok self critique assessment: 1
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23:25:53 `questionNumber 20000 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> vol= area x h = 50cm^2 x 60 cm =3000 cm^3 confidence assessment: 2
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23:27:26 09-14-2008 23:27:26 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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NOTES -------> We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box. So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.
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23:30:22 `questionNumber 20000 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> vol= 1/3 area x h = 1/3 ( 20 m^2 x 9m) =60 m^3 confidence assessment: 2
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23:31:22 `questionNumber 20000 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone. In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.
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RESPONSE --> Thank goodness! self critique assessment:
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23:35:35 `questionNumber 20000 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> vol= 4/3 (pi x r^3) = 4/3 ( 3.14 x 64m^3) 4/3 x 200.96 m^3 = 267.95 m^3 confidence assessment: 2
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23:36:47 `questionNumber 20000 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
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RESPONSE --> ok self critique assessment:
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23:46:11 `questionNumber 20000 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> v= 4/3 (3.14 x r^2), diam = 14000 km so, r=7000km = 4/3 (3.14 x (7,000km)^3 = 4/3 x 3.43 x 10^11 = 4.57 x 10 ^11 confidence assessment: 2
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23:51:51 `questionNumber 20000 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> ok self critique assessment: 1
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23:52:27 `questionNumber 20000 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> volume= area x height confidence assessment: 3
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23:52:47 `questionNumber 20000 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> ok self critique assessment: 3
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23:53:24 `questionNumber 20000 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> volume = 1/3 area x height confidence assessment: 3
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23:53:38 `questionNumber 20000 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> ok self critique assessment: 3
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23:54:17 `questionNumber 20000 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> volume = 4/3 x pi r^3 confidence assessment: 3
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23:54:32 `questionNumber 20000 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> ok self critique assessment: 3
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23:57:30 `questionNumber 20000 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have organized my knowledge by following the examples and working the problems in the order that they are given. confidence assessment: 2
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NƆّXޝ assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 09-15-2008
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00:02:49 `questionNumber 30000 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> sa= 3m x 4m x 6m= = 72m^3 confidence assessment: 1
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00:07:42 `questionNumber 30000 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
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RESPONSE --> now I get it. Didn't know. self critique assessment: 1
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00:14:11 `questionNumber 30000 `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
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RESPONSE --> I have no idea. confidence assessment: 0
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00:20:03 `questionNumber 30000 The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.
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RESPONSE --> ok confidence assessment: 1
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00:24:56 `questionNumber 30000 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> area= 4(pi) (r^2) = 4 (pi)(1.5)^2 = 4 x pi x 2.25 =9 pi cm^2 confidence assessment: 2
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00:25:13 `questionNumber 30000 The surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.
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RESPONSE --> ok self critique assessment: 2
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00:28:39 `questionNumber 30000 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> c^2= a^2 + b^2 = 5^2 + 9^2 = 25 + 81 =10.3 m confidence assessment: 2
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00:29:12 `questionNumber 30000 The Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.
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RESPONSE --> self critique assessment: 2
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00:31:46 `questionNumber 30000 `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?
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RESPONSE --> 6m - 4m = b^2 = 2m = 1.4m confidence assessment: 1
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00:33:58 `questionNumber 30000 If c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m.
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RESPONSE --> I forgot to square ^2 my answer when I solved for the unknown. I have it now self critique assessment: 2
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00:44:44 `questionNumber 30000 `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?
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RESPONSE --> d= m g/vol. cm^3 = 700 g/ (4 x 7 x 12) = 700/ 336 = 2.08 G/CM ^3 confidence assessment: 3
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00:45:18 `questionNumber 30000 The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).
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RESPONSE --> self critique assessment: 2
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00:52:18 `questionNumber 30000 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
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RESPONSE --> D= M/ V 4/3 X PI x 4^3 = vol. 4/3 x 3.14 x 64= 267.95cm ^3 3000/267.95=11.20kg confidence assessment: 1
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00:54:35 `questionNumber 30000 A average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> ok self critique assessment:
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00:58:06 `questionNumber 30000 `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?
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RESPONSE --> 3g/cm^3 confidence assessment: 1
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00:59:30 `questionNumber 30000 The first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.
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RESPONSE --> ok self critique assessment: 2
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01:01:23 `questionNumber 30000 `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?
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RESPONSE --> confidence assessment: 0
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01:02:35 `questionNumber 30000 We find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..
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RESPONSE --> self critique assessment: 0
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01:04:10 09-15-2008 01:04:10 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
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NOTES ------->
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01:05:54 `questionNumber 30000 The volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information.
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RESPONSE --> I tried to click save as notes to give answer, but it went to the next one. self critique assessment: 0
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01:07:49 `questionNumber 30000 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> Find the circumfeence, The area of the curved side = circ. x altitude Find area, then total area. confidence assessment: 3
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01:08:13 09-15-2008 01:08:13 The curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2.
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01:08:53 `questionNumber 30000 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> A= 4 pi r^2 confidence assessment: 2
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01:09:05 `questionNumber 30000 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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RESPONSE --> self critique assessment: 2
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01:10:01 `questionNumber 30000 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> density= mass/volume g/cm^3 confidence assessment: 3
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01:10:24 `questionNumber 30000 The average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'
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RESPONSE --> self critique assessment: 2
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01:10:57 `questionNumber 30000 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> volume= area x height confidence assessment: 2
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01:11:37 `questionNumber 30000 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> self critique assessment: 1
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01:12:37 `questionNumber 30000 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have tried to organize my knowledge in a manner that I can understand for hopeful future use. confidence assessment: 2
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