course Phy 121 Please let me know when you get this and if there is anything I am lacking for this first set of assignments. I just do not want to be penalized for not having something I have completed. x̰IwȐassignment #001
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17:13:13 `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 = 2L * 2W = (2*4) + (2*3) = 8+6 = 14 square meters confidence assessment: 1
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17:17:10 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 --> I remembered the incorrect formula for area. It should have been length times width instead of twice the width and length. The part of the solution about division of rows containing squares makes sense, but that's not the first thing I thought of so I didn't include it in my answer. I understand the concept of meters squared. self critique assessment: 2
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17:21:50 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> The formula for the area of a triangle is 1/2hb. Therefore, A = (1/2)(4.0)(3.0)= 6 m^2 confidence assessment: 1
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17:23:58 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 did not think of considering this triangle as part of a rectangle. I looked up the formula for the area of a triangle. I achieved the correct answer. self critique assessment: 3
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17:25:26 `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 --> The area of a parallelogram is A = bh. Therefore, A = (5.0)(2.0) = 10.0 m^2 confidence assessment: 1
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17:27:22 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 --> Once again the rectangle is mentioned in the solution. I understand that it is easier for most people to think of things in terms of rectangles rather than to memorize a bunch of formulas. However, it is easier for me to work things in terms of using a formula. I achieved the correct answer. self critique assessment: 2
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17:29:13 `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 --> The formula for the area of a triangle is: A = 1/2 hb A = (1/2)(2.0)(5.0) = 5 cm^2 confidence assessment: 2
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17:30:18 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 --> I need to pay closer attention to significant digits, but I achieved the correct answer. self critique assessment: 2
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17:32:20 `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 --> In this case A = bh. A = (4.0)(5.0) = 20.0 km^2 confidence assessment: 1
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17:35:25 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 --> I was unaware (or failed to remember) the concept of average altitude concerning the trapezoid, and I can remember being taught in grade school that several figures can be rearranged to form triangles. But, once again it is easier for me by plugging into formulas. I did achieve the correct answer. self critique assessment: 2
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17:38:02 `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 --> The formula for the area of a trapezoid is: A = bh A = (4)(3.0)(8.0) = 96 cm^2 confidence assessment: 1
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17:39:24 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> I had forgotten the concept of average altitude. I mistook width for height, and that's where I made my mistake. self critique assessment: 2
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17:41:23 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> The formula for the area of a circle is: A = (pi)r^2 = (pi)(3.00^2) = 28.27 cm^2 confidence assessment: 1
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17:44:24 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 --> I got the right number, but I used the incorrect units and incorrect number of significant digits. The reminder about not confusing things was good to know. self critique assessment: 2
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17:51:31 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> Twice the radius will give you diameter. So the diameter of the circle is 6 cm. The formula for Circumference of a circle is: C = (pi)d = 2 (pi) r C = (pi)6 = 18.8 cm^2 C = 2 (pi) 3 = 18.8 cm^2 confidence assessment: 1
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17:56:06 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 --> If the correct answer is indeed 18.84, then I got the correct number and put down the incorrect units. I guess since the unit answer for most of the other problems so far has been cm^2 I got in the habit of putting it down and most likely that is where I made my mistake. Or if you could please explain to me again the how to end up with a units squared answer. self critique assessment: 2
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17:57:44 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> The formula for the area of a circle is: C = (pi)d C = (pi) 12 = 37.7 meters confidence assessment: 1
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17:58:53 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.
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RESPONSE --> This was a stupid mistake made on my part. I used the wrong formula and I understand the solution. self critique assessment: 2
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18:08:16 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> The best way I know to do this is to disect the formula to understand how I know the information given in the problem. Our circumference is 14'pi meters. The formula for circumference is: C = (pi)d Working backwards we can used simple algebra to detemine that the diameter is 4.46. Area of a circle can also be found using the formula: A = (pi)d^2 / 4 = (pi)(4.46^2) / 4 = 15.6 pi m^2 confidence assessment: 1
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18:11:15 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 --> I think I made my mistake mostly through working backwards trying to find diameter instead of radius because I understand how the solution was obtained. self critique assessment: 2
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18:16:39 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> The formula for the area of a circle is: A = (pi)r^2 I will plug in what I know at this point: 78 square meters = (3.14) r^2 Then divide everything by 3.14: 24.8 = r^2 Take the square root of both sides: 4.98 = r confidence assessment: 1
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18:18:07 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 --> My rounding wasn't exactly as is listed here, but I understand the process of solving the problem. I also left off the units. self critique assessment: 2
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18:20:13 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> For example, if we have a 4 cm by 3 cm rectangle there was previous mention of visualizing 3 rows with 4 squares in each row. confidence assessment: 1
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18:21:20 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 didn't mention the formula, but now I know how much information to include in my answers. self critique assessment: 2
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18:24:18 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> The area of a right triangle is: A = 1/2 hb We can visualize the right angle's hypotenuse as the same length of the hypotenuse of another right triangle which can be used to form the other half of a rectangle. Leading us back to our A = L * W. confidence assessment: 1
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18:24:48 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 --> I think I explained the same points that are mentioned here. self critique assessment: 2
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18:25:33 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> To calculate the area of a parallelogram we can use the following formula: A = bh confidence assessment: 1
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18:26:12 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 --> I got the correct answer. self critique assessment: 3
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18:28:29 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> To calculate the area of a trapezoid we can use the formula: A = average altitude * base confidence assessment: 1
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18:29:00 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 --> I said height instead of width. self critique assessment: 2
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18:30:26 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> We can calculate the area of a circle by using the formula: A = (pi) r^2 confidence assessment: 1
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18:30:43 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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18:37:42 `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 --> We can calculate the circumference by using the formula: C = 2 (pi) r The area for a circle is: A = (pi) r^2 The order of operations is different for each formula. With the C formula we can multiply 2 times pi times r. With the A formula the radius is squared and then multiplied by pi. In grade school I was told to remember that pi are not round, pi are squared. confidence assessment: 1
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18:38:17 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 --> My response was mostly correct. I forgot to mention that circumference is measured in squared units. self critique assessment: 2
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18:42:23 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> These problems served as a good review of algebraic principles of problem solving and determining area and circumference. I am now better able to remember different ways of approaching a problem, and I have been reminded of the way I should try to be thinking when solving these problems. confidence assessment: 1
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18:42:46 This ends the first assignment.
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RESPONSE --> I have reached the end of the first assignment. confidence assessment: 2
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N{yh{s assignment #002 002. Volumes qa areas volumes misc 06-02-2007
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19:53:00 `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 --> The formula for calculating the volume of a rectangular solid is: V = lwh V = 3*5*7 = 105 cm^3 confidence assessment: 1
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19:57:59 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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RESPONSE --> This solution reminds me of the formula V = A*h and of how that particular concept works; however, I still achieved the correct answer. self critique assessment: 2
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20:00:27 `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 --> If applying the formula V = A*h, then we can plug in the following: V = 48*2 = 96 m^3 confidence assessment: 1
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20:01:10 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 --> I got the correct answer. self critique assessment: 2
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20:04:26 `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 --> Once again, I will apply the formula V = A*h = 20*40 = 800 m^3 confidence assessment: 1
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20:06:02 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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RESPONSE --> I chose to apply the formula V = A*h because it stated in a previous solution that this formula applied to a broader range of objects. Therefore, due to the information given I also applied it here which resulted in the correct answer. self critique assessment: 2
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20:13:14 `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 --> The formula for finding the volume of a cylinder is: V = (pi)r^2 l Length is not given here but since the altitude is 30 cm, I will plug this in for length. V = (pi) 5^2 (30) = 2356 cm^3 confidence assessment: 1
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20:15:26 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 --> I made my mistake by not using the formula for the area of a circle, and by assuming 30 cm altitude for length. I thought the V = A*h applied somewhere, but I wasn't sure how to use it. self critique assessment: 2
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20:29:02 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> I measured a can of Chicken Noodle Soup. 2 1/2 inches across top, 4 inches long. V = A*h and A = (pi)r^2 Since 2.5 is the diameter of the top of the can the radius is 1.25 inches. A = (pi) 1.25^2 = 4.9 square inches V = 4.9 * 4 = 19.6 inches^3 confidence assessment: 1
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20:30:22 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 --> As best I can tell from reading the solution, I did this problem correctly. self critique assessment: 2
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20:31:49 `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 --> Once again, I will apply the formula V = A*h: V = 50*60 = 3000 cm^3 confidence assessment: 1
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20:33:07 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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RESPONSE --> If I had taken 1/3 of 3000, to take into account the pyramid structure I would have ended up with the correct answer. self critique assessment: 2
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20:36:05 `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 --> I will apply the formula V = A*h and take 1/3 of the answer to account for the fact that the structure is a cone. V = 20*9 = 180 m^3 1/3 of 180 = 60 m^3 confidence assessment: 1
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20:36:33 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 --> I got the correct answer. self critique assessment: 3
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20:41:38 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> According to the textbook, the volume of a sphere is: V = 4/3 (pi) r^3 V = 4/3 (pi) 4^3 = 268 m^3 confidence assessment: 1
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20:46:58 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 --> I used the same formula and got a different answer which is undoubtedly a difference in the order of operations, but 4^3 = 64. 4/3 pi * 64. 64/1 * 4/3 = 256/3 pi m^3. I see where I made my mistake. self critique assessment: 2
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20:51:50 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> Diameter divided by 2 gives us the radius which is 7000 km. The formula for finding the volume of a sphere is: V = 4/3 pi r^3 V = 4/3 pi 7000^3 = 4/3 pi 3.43*e11 = (1.372*e12)/3 pi km^3 confidence assessment: 1
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20:52:16 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 --> I got the correct answer. self critique assessment: 3
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20:53:41 `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 --> The formula V = A*h would be applied the the units would be cubic (^3). confidence assessment: 1
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20:54:24 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 --> Although I did not explain what each variable means, I got the correct formula and units. self critique assessment: 2
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20:55:17 `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 --> The formula V = A*h would be applied, and then you would multiply the resulting answer by 1/3. confidence assessment: 1
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20:55:35 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 --> I got the correct answer. self critique assessment: 3
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20:56:22 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> The formula for the volume of a sphere is: V = 4/3 pi r^3 confidence assessment: 1
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20:56:38 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 --> I got the correct answer. self critique assessment: 3
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20:59:21 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> This set of problems served as a great review on how to find volumes and areas of various objects. I particularly appreciated my new (or just forgotten) knowledge of the formula V = A*h since it can be applied to a broader range of objects. This set of exercises has also improved my way of thinking when approaching these types of problems because I think I can see improvement from one exercise to the next. confidence assessment: 1
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20:59:42 This ends the second assignment.
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RESPONSE --> I have reached the end of the second assignment. confidence assessment: 2
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ؗ·yۜ|wI~ assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 06-02-2007
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21:12:59 `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 --> I'm thinking the surface area of a rectangle is determined by side 1 + side 2, etc. since volume is determined by length times width times height. I guess I will know for sure shortly. 3 + 4 + 6 = 13 m confidence assessment: 0
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21:14:34 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 --> I could not remember how to do this sort of problem or how to set it up properly. Now that I have seen the solution, I'm sure I would be able to do one like it without trouble. self critique assessment: 2
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21:27:23 `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'm really not sure at all about how to approach this problem. The textbook formula for the surface area of a cylinder is: V = 2 pi r l + 2 pi r^2 I will plug in altitude for length since I don't know what else to do. V = 2 pi 5 12+ 2 pi 5*2 = 2 pi 60 + 2 pi 25 = 534 pi m^3 confidence assessment: 0
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21:32:13 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 --> This problem definately served as a learning experience for me. I probably knew how to do this at some point and have forgotten over time. But, I didn't think about circumference at all. If I had had some sort of example to go on, I would have had a better chance of getting this problem right. But, know that I know how to set it up I would have less trouble working a similar problem. confidence assessment: 2
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21:34:50 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> The formula for the surface area of a sphere, according to the textbook, is: V = 4 pi r^2 Division of the diameter by 2 gives me the radius which is 1.5. V = 4 pi 1.5^2 = 4 pi 2.25 = 28 pi m^2 confidence assessment: 0
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21:37:42 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 --> I used the correct formula, but why would 3/2 squared not then be multiplied by 4 pi? self critique assessment: 2
Also it 's generally easier to connect a number like 25 pi or 9 pi to the information given in the problem. You can recognize 25 as the square of 5, whereas you probably won't recognize any connection between 78.5 (which is 25 pi) and 5.
However when the information given is approximate in nature, or when an approximate solution fits the situation, it's fine to multiply the quantities out. &#
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21:47:25 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> I will attempt to get this answer using the pythagorean solution. a^2 + b^2 = c^2 5^2 + 9^2 = c^2 25 + 81= c^2 106 = c^2 10.2 m = c confidence assessment: 1
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21:48:17 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 --> It looks as though I got the correct answer. self critique assessment: 2
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21:51:55 `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 --> Once again I will use the pythagorean theorem to find the answer. a^2 + b^2 = c^2 4^2 + b^2 = 6^2 16 + b^2 = 36 b^2 = 36 - 16 = 20 b = 4.5 m confidence assessment: 1
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21:52:11 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 got the correct answer. self critique assessment: 3
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21:55:40 `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 --> Unless I'm mistaken, I'm pretty sure the formula for density is D = mass/volume. To find the volume of the rectangular solid we first need to use the formula V = lwh. V = 4*7*12 = 336 cm^3 D = 700/336 = 2.1 g/cm^3 confidence assessment: 1
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21:56:51 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 --> I got the correct answer and I understand the solution presented here. self critique assessment: 2
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22:02:29 `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 --> The formula for density is: D = m/V We know the density, but we don't know mass or volume, however we can find the volume of the sphere because we have been given its radius. V = 4/3 pi r^3 V = 4/3 pi 4^3 = 4/3 pi 64/27 = 256/81 pi m^3 Now that we know D and V we can find m by rearranging the formula. D*V =m 3000 * 256/81 = 9481 kg = m confidence assessment: 1
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22:04:24 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 --> Considering the fact that I did not take into account average density along with some other mathematical error, contributed mostly to my incorrect answer. However, I do mostly understand the process necessary to solve the problem. self critique assessment: 2
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22:08:00 `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 --> The difference in the change is 4 grams, and difference in the change in density is 4 cm^3. Every way I calculate this its looks as though the average density is 4 g / cm^3. confidence assessment: 0
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22:09:47 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 --> My mistake here was that I calculated all the difference mentioned in the solution, but then I didn't know what to do with them as far as plugging them into this final formula. self critique assessment: 2
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22:16:12 `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 --> We need these formulas to solve this problem: V = lwh D = m/V Average Density = Total mass / total volume V = 2*3*5 = 30 m^3 Now we need to determine the mass of the sand and the cannon balls. Sand: 2100 = m / 30 m= 63000 Cannon balls: 8000 = m / 30 m = 240000 AD = (63000+240000) / 30 = 10100 kg / m^3 confidence assessment: 0
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22:17:14 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 --> Aside from incorrectly calculating the mass of the sand it looks like I did everything else right. self critique assessment: 2
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22:28:05 `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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RESPONSE --> The formula for density is: D = m/V 860 = m/V We also need to find volume: V = lwh or V = bh V = 1,700,000 * .015 = 25500 m^3 860 = m / 25500 860 * 25500 = 21,930,000 kg = m confidence assessment: 0
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22:29:17 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 --> My answer is close to the correct one, and would have been exactly right had I used V = A*h. self critique assessment: 2
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22:29:57 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> The area of a cylinder is found by using: A = circumference * altitude confidence assessment: 1
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22:30:43 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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RESPONSE --> My answer encompassed part of what was mentioned here. self critique assessment: 2
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22:31:51 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The formula for the surface area of a sphere is: A = 4 pi r^2 confidence assessment: 1
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22:32:04 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 --> I got the correct answer. self critique assessment: 3
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22:33:13 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is defined as mass per unit of volume. D = m/V confidence assessment: 1
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22:33:59 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 --> I got the correct answer. self critique assessment: 3
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22:37:07 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Since the formula for density is: D = m/V a simple rearrangment of the formula will help us find volume. Multiply each side by V to get: m = V*D, then divide each side by D to get: m/D = V confidence assessment: 1
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22:37:30 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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22:38:12 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I had a really difficult time with this particular assignment, any additional comments or suggestions you can provide would be most appreciated. confidence assessment: 1
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22:38:29 This ends the third assignment.
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RESPONSE --> I have reached the end of the third assignment. confidence assessment: 3
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T_ؕͪefD assignment #001 001. Rates qa rates 06-02-2007
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10:50:20 `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'.
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RESPONSE --> I have read and I understand the instructions. confidence assessment: 2
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ppoaV˦ assignment #001 001. Rates qa rates 06-02-2007
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10:58:31 `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'.
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RESPONSE --> Copy and paste these instructions elsewhere, so I can refer to them later. For the next question or answer, I am to click on the button in the top left hand corner of this window that says ""Next Question / Answer"". I must answer the question before I click on this button again. Do not mess with information presented in the left hand box. If I should get an incorrect answer, I need to do a self-critique. If I see where I made my mistake and learned something from it, I need to state this in the self critique to reinforce learning. Make notes as appropriate, and don't copy/paste anything from the left hand box because that information is already saved. My response can be 'Saved as Notes'; if I choose to do that I can choose that option instead of Enter Response. After choosing one or the other of the previously mentioned options, I am to click on the Next Question/Answer button to proceed. confidence assessment: 2
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10:58:45 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
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RESPONSE --> Ok confidence assessment: 2
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11:00:55 `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?
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RESPONSE --> You are earning money at a rate of $10.00 per hour. $50 / 5 hours = $10.00 confidence assessment: 3
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11:24:49 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.
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RESPONSE --> I got the correct answer, and I undrestand the concept of pay rate. confidence assessment: 2
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11:28:25 `q003.If you make $60,000 per year then how much do you make per month?
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RESPONSE --> You will make $5,000.00 per month. $60,000.00 per year divided by 12 months = $5,000.00 per month. confidence assessment: 3
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11:29:09 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.
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RESPONSE --> I have understood everything encountered in the problems so far. confidence assessment: 3
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11:35:13 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
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RESPONSE --> It would be more accurate to say that the business makes an average of $5000 per month. It is unlikely that a business will make exactly $5000 each month, it is actually more probable that the business will make a little more or a little less than that amount each month resulting in an average of $5000 per month. confidence assessment: 2
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11:40:14 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.
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RESPONSE --> When answering this question I did not specifically mention seasonal and other market fluctuations. I also failed to mention demand for services or commodities provided by the business. I did not consider all the things business must deal with concerning finances. I took more of a mathematical approach to answering the problem and considered the probability of the business making the exact same amount of money each month for a year. confidence assessment: 3
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11:44:30 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
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RESPONSE --> You would be covering distance at a rate of 50 miles per hour. 300 miles / 6 hours = 50 miles per hour We say average rate instead of just plain rate because it is unlikely that the mode of transportation will go exactly 50 miles per hour. For example, if you are driving a car to cover this distance (unless you set the cruise control) it is unlikely that your foot is going to hold the accelerator at exactly 50 miles per hour during the entire trip. You must also take into account stopping for red lights, people who turn off in front of you, etc. confidence assessment: 2
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11:50:23 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.
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RESPONSE --> I did not include the concept of speed in my answer, however, I do understand this concept now it has been mentioned in the solution. I was happy with the fact that the solution contained mention of a car as an example, but I also did not specifically state 'slight changes in pressure on the accelerator' in my answer. But, I mention that someone's foot is unlikely to hold the accelerator at exactly 50 miles per hour for the entire trip, so I at least partially covered the previously mentioned specific statement. confidence assessment: 2
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11:58:52 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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RESPONSE --> You would be using 1 gallon of gasoline for every 20 miles traveled. 1200 miles / 60 gallons = 20 miles per gallon 20 miles * 60 gallons of gasoline = you can travel a distance of 1200 miles. confidence assessment: 2
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12:09:03 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.
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RESPONSE --> I think I was on the right track with my answer and expressed it in wrong terms. I had forgotten change in first divided by change in the second, but I did know that division was necessary. I also expressed my answer in terms of miles per gallon, which was not what the question asked for. It looks as though I did the division backwards, I divided 1200 / 60 instead of 60 / 1200. After reading the solution I understand the difference in terms of answer, and I see the mistake I made. confidence assessment: 2
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12:15:50 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
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RESPONSE --> It isn't necessary to add because we have totals. Allow me to explain: In one of the earlier questions we were asked to calculate rate of pay per hour. $50 per hour / 5 hours = $10 per hour. to check that answer we would be required to add. $10 (hour 1) + $10 (hour 2) + $10 (hour 3) + $10 (hour 4) + $10 (hour 5) = $50 for 5 hours of work. If you have a total number of hours, and the total of money earned the average is determined by division to obtain the average number of dollars earned per hour. confidence assessment: 2
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12:21:08 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.
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RESPONSE --> The solution presented here is a better way of saying what I was trying to get at. I knew why we didn't have to add, but I didn't do too well at trying to explain it. I did know that we were expecting to earn different amounts (concerning different questions), but I failed to mention this in my answer. Knowing the individual quantities should have resulted in the same average calculated (as I have been doing) in previous problems. I understand the concept presented here, but could not explain it properly in my previous answer. confidence assessment: 2
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12:46:18 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
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RESPONSE --> First of all, I am not sure what units in which to quantify lifting strength; but, I think I can still make my point. We are given the information that the first group did 10 PU / 1 day for a year and group two did 50 PU / 1 day for a year. A year is made up of 365 days. Therefore, in a years' time group 1 did 3650 pushups and group 2 did 18250 pushups. So group 2 did five times as many pushups as the first group. (18250 / 3650 = 5) If you divide 365 / 10 = 36.5. This number is not the result we are looking for. However, if 10/365 = .027. I am pretty sure this number is the rate of strength increase per pushup per day. So, over the course of one day the completion of 10 pushups results in a change of .27 in lifting strength. The same applies when considering 50 pushups. 50/365 = .137 .137 *50 = 6.85 change in lifting strength per day I go back to the numbers .027 and .137; if added and divided by two the result is .082. Because the question does not ask for rate of change in lifting strength per day per group, I am assuming I must find this for both group together. Obviously, lifting strength increasing per day per group will vary because one group is doing more pushups than the other. So, I am going to say that average rate of lifting strength per daily pushup for both groups is approximately .082. confidence assessment: 1
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12:52:22 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.
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RESPONSE --> It looks like I made this problem way more complicated than it had to be. I completely missed that I need to look at differences between pounds and then between number of pushups to do divisions (as done in previous problems) to obtain the correct answer. I really wasn't sure how to set up this particular problem to start with, so I went through several calculations hoping that I would figure it out. Now that I have read the solution the answer makes sense, but I wasn't able to recall this type of process from memory to start with so I have some reviewing to do. confidence assessment: 1
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13:10:24 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
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RESPONSE --> If this problem is answered in a similar manner as the one before it, all groups are doing 30 pushups per day, and there is a 20 pound difference in the weight being used. 20/30 = .67 I would conclude that this is the average rate at which lifting strength increased. confidence assessment: 1
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13:12:01 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.
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RESPONSE --> This time I missed the question because I didn't take into account the difference in lifting strength of 17 pounds. confidence assessment: 2
ave rate = (change in A) / (change in B).
From this it follows that
change in A = ave rate * change in B
and also that
change in B = change in A / ave rate
where 'ave rate' here is an abbreviation for the specific term 'average rate of change of A with respect to B.
On a graph of A vs. B, the quantity A is represented relative to the vertical axis and the quantity B relative to the horizontal. Between two graph points, therefore, the change in A is the 'rise' from one point to the other, and the change in B is the 'run' from the same first point to the second. The average rate of change of A with respect to B is then
ave rate = (change in A) / (change in B) = rise / run.
Since rise / run between two points is the slope of the straight line segment between those points, we can identify an average rate of change of A with respect to B as the slope between two points on the graph of A vs. B.
Since slope = rise / run, we see that
rise = slope * run and
run = rise / slope
Interpreting slope as ave rate, rise as change in A and run as change in B, this again tells us that change in A = ave rate * change in B, and change in B = change in A / ave rate. &#
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13:22:29 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
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RESPONSE --> Initially I would calculate the rate of distance covered between each respective point. 100/12 = 8.3 m/s 200/22 = 9.0 m/s But if the total is considered: It took the runner 34 seconds to cover 200 meters. 200/34 = 5.88 This leads me to conclude that the runner was covering distance at an average rate of 5.8 meters per second. confidence assessment: 2
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13:26:18 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.
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RESPONSE --> Where did the 10 seconds come from? I understand that the runner traveled 100 meters between the two positions, but I guess I missed what the question was asking. I would better understand the solution if I knew where the 10 seconds came from. Everything else seems to make logical sense. confidence assessment: 2
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13:34:52 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
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RESPONSE --> My best estimate of how long it takes the runner takes the runner is 12 seconds. If the 100 meter mark is passed while moving at 10 m/s then it will take 10 seconds for the runner to reach that point. (100/10 = 10) And if the runner reaches the 200 meter mark moving at 9m/s it takes the runner 22 seconds to cover that distance. The difference between 22 and 10 is 12. confidence assessment: 1
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13:41:25 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.
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RESPONSE --> When answering the question I didn't take into account that the runner would slow down. I also assumed average speed, which I should not have done. Had I not made these mistakes I would have achieved the correct response, and I understand the solution presented here. confidence assessment: 2
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13:43:11 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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RESPONSE --> We do it now because we do not know average speed, but we do have individual quantities. So, we are able to calculate an average speed so we can answer other question pertaining to a particular situation. confidence assessment: 1
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13:45:53 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.
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RESPONSE --> I think with my answer I was getting at the same point, but I explained it differently. I did not specifically mention the word rate, but I did mention that we had individual quantities to work with so we could find average rate. My problem here is that I could have explained it better. confidence assessment: 2
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13:46:08 end program
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RESPONSE --> I reached the end of the program. confidence assessment: 2
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