course Phy 122 This ends the third assignment. ????????????assignment #001
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11:21:26 `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 --> 4m x 3m = 12m^2 is the area of a rectangle confidence assessment: 3
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11:21:45 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 --> OK self critique assessment: 3
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11:23:23 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> 4m x 4m x 3m = 48m^2. How do you find the area of a right triangle. confidence assessment: 1
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11:26:54 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 --> Ok now I understand that in order to find the area of a right triangle without the base or altitude then you have to make a rectangle out of 2 of the right triangles. This makes one of the legs of the triangle one side of the rectangle and the other leg the other rectangle side. Then find the area L x W and then divide by 2 since there are two triangles. self critique assessment: 2
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11:30:10 `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 --> I am not quite sure how to do this equation, although it seems to be similar to the last just backwards. My guess would be treating it like a rectangle and doing 5m x 2m = 10m confidence assessment: 0
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11:31:02 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h. The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.
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RESPONSE --> Ok I was right although I should have squared the resulting m to show area. self critique assessment: 2
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11:33:08 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> A = 1/2 + b x h 1/2 + 5cm x 2cm = 5cm^2 How do you find the area of a triangle using altitude? confidence assessment: 1
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11:35:35 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 --> Ok, so although I was not sure of my answer, I was right with the right reasoning. I can make a parallelogram like done in the previous question and find the area of one triangle by taking half. I just realized that I made the mistake of saying 1/2 + instead of x but that was just a typing error. self critique assessment: 2
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11:36:18 `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 --> I do not even remember what a trapezoid is as sad as that sounds. confidence assessment: 0
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11:37:30 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> Ok, so it is the same equation we have been using for rectangles applies to trapezoids also. self critique assessment: 2
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11:39:21 `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 --> First I would take the average of the altitudes which is 3cm + 8cm = 11cm / 2 = 5.5cm then multiply by 4cm like I would a rectangle to get 22 cm^2 confidence assessment: 3
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11:39:44 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> Ok, I understand it now. self critique assessment: 3
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11:40:56 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> I do not remember how to find the area of a circle. All I remember is something to do with pie and that the radius is half the diameter. confidence assessment: 0
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11:42:56 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 so I have to remember that the equation is A= pi*r^2 self critique assessment: 3
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11:45:18 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C = 2pi * r which is = 2pi *3cm. In keeping it exact, the result would be left this way because anything else solving for pi would be an approximation. But if done, it would be 18.85 cm confidence assessment: 3
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11:49:54 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 --> I did not see that I could simplify the equation further to 6 pi cm. self critique assessment: 3
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11:54:00 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> half of 12m is 6m which is the radius which then can be used in the equation A= pi*r^2. A= pi*6^2 = pi*36 m^2. confidence assessment: 3
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11:54:20 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.
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RESPONSE --> Ok self critique assessment: 3
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11:55:28 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> I really dont remember confidence assessment: 0
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11:57:55 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 am still not really clear on what is going on in this equation. self critique assessment: 0
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11:58:53 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 --> self critique assessment: 0
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11:59:04 01-16-2007 11:59:04 `q011. What is the radius of circle whose area is 78 square meters?
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NOTES ------->
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11:59:58 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ). Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.
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RESPONSE --> Ok I accidently pressed save as notes when the question popped up and now its just the answer. I don't know what to do here. self critique assessment: 0
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12:01:15 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> Just picture the length and width of the two sides and then multipy them together. confidence assessment: 2
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12:29:36 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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RESPONSE --> Ok, I did not know we were visualizing that deeply. self critique assessment: 3
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12:31:34 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> We visualize the triangle doubled and made into a rectangle, then multiply the blocks in a row times the rows. Then take that and divide by two since there is two triangles. confidence assessment: 3
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12:31:52 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 --> ok self critique assessment: 3
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12:34:48 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Pretty much the same way we calculate a rectangle and the triangles. A= 1/2*L*W. I am not sure if I remembered this correctly confidence assessment: 2
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12:35:22 01-16-2007 12:35:22 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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NOTES -------> Ok now I understand.
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12:36:21 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> Of course I did not write this down thinking that I would remember it. That was a mistake confidence assessment: 0
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12:36:32 01-16-2007 12:36:32 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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NOTES ------->
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12:36:35 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 --> self critique assessment: 3
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12:38:03 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> We take the radius and square it. Then we multiply by pi. ie A=pi*r^2. confidence assessment: 3
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12:38:14 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok self critique assessment: 3
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12:40:26 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> C= 2*pi*r where r is the radius. The circumference is 2 dimensional and is not in a unit squared whereas the area is 3 dimensional. confidence assessment: 2
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12:40:48 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> ok self critique assessment: 3
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12:43:04 `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 remembered most of them from before, put some of them as notes on this program and wrote some of them down on a note book paper. confidence assessment: 3
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12:43:12 This ends the first assignment.
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RESPONSE --> confidence assessment:
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?}??????y?h??????assignment #002 002. Volumes qa areas volumes misc 01-16-2007
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12:49:55 `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 volume is 3cm * 5cm * 7cm = 105 cm^3 confidence assessment: 3
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12:50:52 01-16-2007 12:50:52 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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NOTES ------->
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12:50:56 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 --> ok self critique assessment: 3
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12:53:02 `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 --> 48m^2 * 2m = 96m^3 is the volume of the rectangle. confidence assessment: 3
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12:53:13 Using the idea that V = A * h we find that the volume of this solid is V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.
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RESPONSE --> ok self critique assessment: 3
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17:00:48 `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 --> Although I am not sure how to find the volume of a uniform cylinder, I would assume it would be 20m^2 * 40m = 800 m^3. confidence assessment: 2
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17:01:13 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 --> ok I was right. self critique assessment: 3
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17:04:32 `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 --> How would I find the volume of a cylinder with a radius and altitude? I would assume it would have something to do with getting the circumference of the circles that constitute the top and bottom. confidence assessment: 0
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17:10:13 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 --> Alright so I would have to find the area of the circle using the radius given and then apply that to the V=A*h or multiply the area that was found by the altitude that was given. self critique assessment: 1
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17:15:37 `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 would guess that the altitude would be about 5 inches and the radius being about 1 inch. That would make A= pi * 1in^2 = 2pi * 5 = 10pi in^3 or about 31.42 in^3 confidence assessment: 3
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17:16:49 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using. A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.
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RESPONSE --> Ok I understand self critique assessment: 3
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17:19:25 `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 --> I am not sure how to find the volume of a pyramid but I am just going to guess 50 cm^2 * 60cm = 3,000 cm^3 confidence assessment: 0
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17:23:05 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 --> I would not have guessed that the pyramid would fill a 1/3 of the volume of a box with the V= 1/3 * A * h self critique assessment: 2
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17:24:36 `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 am again not sure on how to find the volume of a cone. confidence assessment: 0
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17:25:53 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 --> Ok so it is the basic concept as applied to the pyramid accept for the cone is applied to the cylinder. self critique assessment: 2
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17:30:54 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> Again, I am not sure of how to find the volume of a sphere. confidence assessment: 0
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17:31:28 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.
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RESPONSE --> ok, now I understand. self critique assessment: 1
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17:35:50 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> Since the diameter is around the outside of the planet and that happens to the a r^3 I can put it into the equation V= 4/3 pi 14,000km = 58643.06287 would be the estimate when solving with pi. Is the diameter r^3? confidence assessment: 1
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17:37:33 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 --> Alright so I was not sure on how the radius could be found from the diameter but now I know that the radius is half of the diameter which then can be placed into the equation. self critique assessment: 2
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17:40:58 `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 --> We either use the area of a circle= pi r^2 times the altitude that is given or put it together to find the volume = pi r^2 h confidence assessment: 3
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17:41:29 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> ok self critique assessment: 3
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17:43:22 `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 cone fills 1/3 of a cylinder and the pyramid is also 1/3 of a rectangle which makes the equation V= 1/3 A *h confidence assessment: 3
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17:43:37 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> ok self critique assessment: 3
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17:44:07 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> V= 4/3 pi r^3 confidence assessment: 3
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17:44:16 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 --> alrighty self critique assessment: 3
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17:45:34 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I remembered some of them and then organized the rest I did not know by looking at the answer and writing down the equation on a notebook paper. confidence assessment: 3
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18:36:02 `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 --> To find the surface area first find the area of one side which would be 4m * 6m= 24m^2 then multiply that by 4 which is the amount of sides, = 96m^2 then find the ends of the rectangle by taking 4m * 3m =12m^2 * 2 = 24m^2. The total result would be 24 + 96 = 2,304 m^2. How do you find the surface area of a rectangular solid? confidence assessment: 1
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18:43:30 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 --> Ok so you just take the dimensions of the pair of three sides, find the area of the of the sides and then multiply by 2 because there are six faces and then add them together to get the total surface area. self critique assessment: 2
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18:45:33 `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 am really slow at this and am not sure of how to do this either. confidence assessment: 0
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18:55:28 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 --> The circumference should first be found using the radius that was given. Then find the area of the sides by using the equation A= C*altitude. Next find the area of the base A= pi r^2. Lastly to find the total area add the area of the sides to 2* the base. confidence assessment: 3
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19:00:01 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> Surface area of a sphere is 4pi r^2 which in this case would be 4pi 3cm^2 = 113.097cm when solving with pi. confidence assessment: 3
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19:03:01 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 may have not remembered the number correctly but why would the radius be divided by 2?
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19:07:21 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> a^2+b^2=c^2 where 5^2 + 9^2 = 106m= c^2 square root of 106= 10.296m=c Is this the right way of finding the hypotenuse of a right triangle? confidence assessment: 1
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19:08:11 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 --> ok, I understand I got this right. self critique assessment: 3
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19:10:45 `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 --> 6^2 + 4^2 = c^2, 52=c^2 square root of 52 is 7.21m confidence assessment: 3
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19:13:52 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 was not paying attention to the fact that this time I was given the hypotenuse and I would have to change the equation so that I was solving for a now and not c. This is a^2 = c^2 - b^2. self critique assessment: 1
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19:19:04 `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 --> Guessing by the density is in grams per cubic cm, I would think that I would first multiply the dimensions = 4cm * 7cm * 12cm = 336cm^3. Then take the mass and divide it by 336cm^3 which is 2.083 g/cm^3. confidence assessment: 1
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19:20:30 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 do not understand why the grams are 2.06 when I did the same thing and I got 2.0833333. Besides that I understand the rest. self critique assessment: 3
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19:25:42 `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 --> I have no idea of how to do this either. confidence assessment: 0
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19:28:51 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 --> All that needs to be done here is to first find the volume of the sphere using the radius given. V=4/3 pi r^3 Then find the mass by taking the density and multiplying by the volume found. self critique assessment: 1
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19:30:01 `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?
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RESPONSE --> Again I feel like I am not prepared for this question. confidence assessment: 0
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19:33:52 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 --> To find the average density first I need to find the mass and add them together to get the total mass. Then to find the average density take the total mass of the two objects and divide by the total volume. self critique assessment: 1
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19:34:42 `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 --> I am completely lost at this point. confidence assessment: 0
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19:37:59 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 --> ok i understand a little now although I think I would need more practice for things like this. self critique assessment: 1
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19:39:58 `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 --> I am still considerably lost here confidence assessment: 0
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19:41:13 The volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information.
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RESPONSE --> I understand it a little better when I see it, but before I have no clue. self critique assessment: 0
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19:43:19 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> We first find the circumference of the cylinder with the given radius. Next find the area of the sides A=c*altitude. Then find the base A= pi r^2. Lastly to find the total area add the sides to 2 times the base. confidence assessment: 3
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19:43:48 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 --> ok I understand self critique assessment: 3
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19:44:48 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The surface area of a sphere is 4 pi r^2 confidence assessment: 2
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19:45:00 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 --> Alright self critique assessment: 3
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19:45:47 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is mass divided by volume of an object. confidence assessment: 3
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19:46:03 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 --> ok self critique assessment: 3
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19:47:46 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Mass divided by density would then equal volume. confidence assessment: 3
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19:48:08 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> ok self critique assessment: 3
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19:49:07 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have wrote the things that I did not remember down on a piece of paper and labeled them things like surface of a rectangle and such followed by the equations. confidence assessment: 3
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