course Phy 121 Again, I apologize for the short answers. I had done this excerise in part 2 of the part III Orientation prior to reading the explinations of self- critique and to include full answers rather than just the answer. |K{ڛE{assignment #001
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12:38:18 `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 --> Yes
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12:38:29 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 --> Yes
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12:39:24 `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 --> The program will never erase my work once I enter responses. Even if the program crashes, as it just did, my answers remain in the SEND file.
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12:39:40 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 --> Got it
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12:40:02 `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 --> Always back up my work. Better safe than sorry.
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|j헮۵Ĺh assignment #001 001. Areas qa areas volumes misc 09-27-2008
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13:04:42 `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 --> 12 m confidence assessment: 3
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13:09:20 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2. The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.
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RESPONSE --> So maybe I did not do the previous problem correctly. self critique assessment: 1
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13:15:20 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> 6.0 m confidence assessment: 1
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13:18:11 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 figured I did not have that one either. self critique assessment: 1
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13:20:28 `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 --> confidence assessment:
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13:21:04 `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 --> 10.0 m ^ 2 confidence assessment: 1
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13:21:41 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 --> Yeah! self critique assessment: 2
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13:22:58 `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 --> 10.0 cm ^ 2 confidence assessment: 1
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13:24:08 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> It figures self critique assessment: 1
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13:30:38 `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 --> 40 km confidence assessment: 1
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13:32:43 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, I feel lost. Is 40 km the same as 20 Km ^ 2? self critique assessment: 1
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13:39:57 `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 --> 28.0 cm confidence assessment: 1
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13:41:43 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> ok, must have got my altitude and width messed up. self critique assessment: 1
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13:55:56 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> 28.26 cm confidence assessment: 1
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13:57:19 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 --> ok square units self critique assessment: 1
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14:14:32 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> 18.84 cm ^ 2 confidence assessment: 1
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14:15:15 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, no ^ 2. self critique assessment: 1
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14:31:55 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> 113.04 cm confidence assessment: 1
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14:32:46 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 --> pretty close self critique assessment: 2
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14:57:14 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> 615.44/4 m confidence assessment: 1
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14:59:01 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 --> umm self critique assessment: 0
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15:10:10 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> .40 m ^ 2 confidence assessment: 0
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15:13:12 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 have no idea what just happened self critique assessment: 0
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15:14:10 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> length multiplied by the width of the rectangle confidence assessment: 3
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15:14:46 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 --> i got the formula anyway self critique assessment: 1
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15:16:02 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> We can join a right triangle to form a rectangle. confidence assessment: 3
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15:16:26 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> yes that self critique assessment: 1
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15:16:58 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Base multipied by height. confidence assessment: 3
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15:17:11 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 --> yes self critique assessment: 2
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15:18:02 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> Form a miror trapezoid so that A = 1/2 b x h confidence assessment: 3
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15:18:37 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 --> oops self critique assessment: 1
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15:19:56 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> A = r x r ^ 2 x pi confidence assessment: 3
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15:20:23 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> yes self critique assessment: 2
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15:20:55 `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?
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RESPONSE --> C = pi x d confidence assessment: 3
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15:21:41 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 --> umm self critique assessment: 1
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15:23:04 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have done the exercises and taken notes. Some of the priciples I still do not understand. confidence assessment: 3
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