course Phy 201 פƇշJ۹
course Phy 201 I am still confused about whether or not I submit more than one assignment per Submit Work form. Please clarify this for me so that I know I am doing the right thing! Thank yo.
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01:18:58 `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=lw A=(4)(3)=12m^2 confidence assessment: 3
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01:19:10 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 --> self critique assessment:
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01:23:09 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> For this, I think one uses the regular triangle area formula which is A=1/2 b(h); after figuring that out, though, since this is a right triangle, one must divide the final answer in half again, because the right triangle is treated like half of another triangle, the long side other than the hypotenuse serving as the ""h"" in the equation. Doing this, I get 6 for the initial equation; then, dividing it by 2 to get the final answer, I get 3. confidence assessment: 2
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01:25:00 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. So you don't divide it by two again--I was close. I will remember next time just to leave the actual triangle formula alone; also, I do think it helps me to remember the formula how you provided the basis for the formula as being derived from a rectangle dissected diagonally. self critique assessment: 2
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01:26:57 `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 think the formula for the area of a parellelogram is merely base x height, which would mean that 5 x 2 would yield an area of 10m^2. confidence assessment: 2
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01:27:14 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 --> I understand this. self critique assessment: 3
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01:29:01 `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 --> Is height the same as altitude? If so, then we just use the 1/2b(h) again. The answer, after filling in 1/2(5)(2) = 5cm^2 confidence assessment: 2
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01:29:15 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 understand. self critique assessment: 3
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09:46:29 `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 think I remember the formula. It is A=1/2a(b1+b2). Unfortunately, I am not sure where to fill in the numbers that this problem gives. My best guess is as follows: A=1/2(5)(4) A=10 confidence assessment: 1
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09:48:14 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 --> This is far easier than actually trying to use the trapezoid formula. One merely needs to know the width and the altitude and one can construct a rectangle and use the A=l(w) formula. self critique assessment: 2
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10:07:58 `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, one should average the altitudes of the trapezoid, giving an altitude average of 5.5. So, we pretend that the trapezoid is like a rectangle and use A=lw, giving us 5.5 x 4 So, the area is 22cm^2 confidence assessment: 2
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10:08:15 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 --> Yay, I understand! self critique assessment: 3
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10:11:27 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> After using A=pi (r)^2, fill in3 for r THe answer we get is 28.27 cm^2. confidence assessment: 3
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10:14:57 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 feel that I understand this well. self critique assessment: 3
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10:16:10 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C=pi(r) So, C=pi (3) = 9.42. confidence assessment: 3
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10:17:56 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 --> Oops, I just put pi (r), not 2 pi r. So, I understand the way to fill in the formula; however, I just, apparently, did not know the circumference formula (sad, sinceyou gave it to me after the last problem!) I will remember now, though! self critique assessment: 2
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10:20:03 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> Ok, take the diameter and divide it by two to give the radius, which is six Then, fill the numbers into the formula: C=2pi(6) =37.7 m confidence assessment: 3
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10:23:25 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 --> I assume, from what you mentioned about signifiacant digits, that that is exactly the reason that my answer varies slightly from yours. self critique assessment: 3
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10:35:55 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> I am going to take a guess. Another formula or way to look at Circumference is to see it as pi (d). D is for diameter. So, we have been told that the Circumference is 14 x pi, meaning that 14 is D. From D, we learn that the radius is 7 (half of the diameter). Then, we merely fill this information into the Area formula, getting pi (7)^2 = 153.9 m^2 confidence assessment: 2
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10:36:48 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 understand this--instead of multiplying out pi in the answer though, you merely include it in the result. I multiplied it out within the equation to elicit the numbers, thus the reason for the answer discrepancies. self critique assessment: 3
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10:38:25 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> First, divide 78 by pi, giving 24.82. Then, we know that 24.82 is equal to r^2. So, we take the square root of both sides, giving r = 4.98 (approximately). confidence assessment: 3
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10:38:52 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 --> Again, same answer; we just rounded differently. I like this kind of math! self critique assessment: 3
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10:40:17 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> I suppose I am not sure what kind of answer this question wants. But I ""visualize,"" I guess, the area of a rectangle as multiplying the measure of the longest side by the measure of the shortest side (hence length x width). It can also be visualized as small cubes that can be counted filling up the entire area. confidence assessment: 3
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10:40:29 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 understand. self critique assessment: 3
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10:43:45 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> We visualize it as half of a rectangle, hence the formula 1/2b(h). confidence assessment: 3
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10:44:02 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 understand. self critique assessment: 3
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10:46:27 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> The area of a parallelogram is base x height, but you can't use the actual sides as height; you must understand that the height must be straight up and down, not angled like the sides. confidence assessment: 3
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10:46:36 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> I understand. self critique assessment: 3
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10:50:29 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> The area of a trapezoid is 1/2(a) x (sum of two bases). confidence assessment: 3
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10:51:05 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 understand. self critique assessment: 3
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10:51:25 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> The area of a circle is pi x radius^2. confidence assessment: 3
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10:51:35 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> I understand. self critique assessment: 3
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10:52:34 `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 --> The circumference is either 2 x pi x r OR pi x diameter. We can keep area versus circumference straight by realizing only area allows for the pi to be squared; with circumference, the pie is always to the first power. confidence assessment: 3
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10:52:43 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 --> I understand. self critique assessment: 3
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10:53:19 `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 taken thorough notes as I have progressed through the section, adding diagrams and examples for each. They can be easily referenced for later use. confidence assessment: 3
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M影ݢWMv assignment #002 002. Volumes qa areas volumes misc 05-14-2007
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11:01:22 `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 --> Volume of a rectangle is length x width x height, so the answer would be 3 x 5 x 7 = 105 cm^3 confidence assessment: 3
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11:02:33 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 --> I understand that area x height lends a greater range of application if one thinks of the formula in those terms. I understand this well. self critique assessment: 3
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11:04:29 `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 --> According to the formula I just learned, one may think of Volume as Area x Height. So, the answer would come from taking 48 m^2 x 2m, yielding the answer of 96m^3. confidence assessment: 3
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11:04:37 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 understand. self critique assessment: 3
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11:22:46 `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 --> V=area x height V=20m^2 x 40m = 800m^3 confidence assessment: 3
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11:22:53 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 understand. self critique assessment: 3
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11:25:21 `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 --> Here is the other formula: V=pi (radius)^2 x height So, V=pi (5)^2 x (30) =2356.2 cm^3 confidence assessment: 3
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11:26:20 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 --> The discrepancy in my answer and yours comes from the fact that I went ahead and multiplied in pi. Either way, I still understand the process. self critique assessment: 3
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11:29:21 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> A metal can has a radius of 2 inches and an altitude of 4. Using the formula and based on my estimates, the volume of the can would be 16pi inches^3 OR 50.3 inches^3 confidence assessment: 3
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11:29:45 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 --> I understand. self critique assessment: 3
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11:32:32 `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 --> V=area x height So, Volume = 50 cm^2 x 60 cm =3000cm^3 confidence assessment: 3
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11:33:51 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 --> uh-oh Since it is 1/3, my answer of 3,000 should have been divided by three, yielding the correct answer. I understand this NOW, and I have made note of it in my notebook. self critique assessment: 2
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11:36:24 `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 think the same ""one-third"" rule applies to this as did to the triangle. So, 20m^3 x 9m = 180 m^3 But then one must divide that by three, yielding the final volume answer of 60 m^3 confidence assessment: 2
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11:36:38 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 understand. self critique assessment: 3
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11:39:35 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> V=4/3 pi x radius^3 V=4/3 pi x (4)63 =268.1m^3 confidence assessment: 2
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11:40:40 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 don't know why our answers are different. Perhaps it is the differences in what number we used for pi. Honestly, I punched my in the calculator, and I do not know to how many places by calculator rounds. Either way, I did the process as you describe to the left; so, I assume my answer is acceptable. self critique assessment: 3
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11:43:41 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> I, of course, used the formula V=4/3 pi r^3 =4/3 pi (7000)^3 = {large number in scientific notation} =1.43x10^12 confidence assessment: 2
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11:44:19 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 understand. self critique assessment: 3
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11:44:58 `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 --> Basically, we take the area of the base of the cylinder multiplied by the height of the cylinder to find the volume. confidence assessment: 3
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11:45:08 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 --> I understand. self critique assessment: 3
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11:46:34 `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 --> One does basically the same thing for a pyramid or cone as one does for a cylinder EXCEPT that after multiplying the area of the base by the height, you divide the number by three. confidence assessment: 3
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11:46:42 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 --> self critique assessment: 3
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11:47:45 `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 4/3 pi x the radius cubed. V=4/3 pi (r)^3 confidence assessment: 3
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11:47:55 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 understand. self critique assessment: 3
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11:48:48 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I have taked very detailed notes in my notebook, including diagrams and examples. I have also made notes of the discrepancies b/w the formulas (i.e., when to divide, in the final step, by three, etc.). I feel that these notes can be easily referenced and utilized in the future. confidence assessment: 3
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k颸դٌ`nx assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 05-14-2007
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11:57:02 `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 --> L=Length W=width and H=Height SA=2LH+2HW+2LW =2(6)(4)+2(4)(3)+2(6)(3) =48+24+36 =108m confidence assessment: 2
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11:57:27 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 --> Uh oh It is meter^2. I didn't put that. Otherwise, I got it. self critique assessment: 2
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12:09:55 `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 --> SA=(2 pi radius x height) + (2 pi radius^2) For only curved sides, we only use the first part of the equation. So, 2xpix(5)x(12) = 377m^2 For the entire surface area, we use the entire equation: We know the first part = 377 So 377 + 2pi(25)= 377+157=534 m^2 confidence assessment: 2
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12:10:34 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 --> Okay, I got the same area once pi is actually multiplied through. confidence assessment: 3
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12:12:40 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> SA=4 pi radius^2 =4pi (1.5)^2 =4 pi (2.25) =9 pi confidence assessment: 3
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12:12:52 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 understand. self critique assessment: 3
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12:17:33 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> For this, we use the pythagoream theorem. x^2 + y^2 = r^2 (5)^2 + (9)^2 = r^2 25+81 = r^2 10.3 = r 10.3 is the length of the hypotenuse confidence assessment: 2
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12:17:46 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 --> I understand. self critique assessment: 3
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12:19:43 `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 --> 4^2 + x^2 = 6^2 16 + x^2 = 36 x^2 = 20 x=4.5 (approx) 4.5 is the length of the ""other"" side confidence assessment: 3
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12:20:04 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 understand. self critique assessment: 3
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12:26:40 `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 --> Density = mass/volume First we have to figure out volume V=area x height =4x7x12=336cm^3 D=m/v =700g/336cm^3=2.08g/cm^3 confidence assessment: 3
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12:27:15 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 understand. self critique assessment: 3
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12:33:20 `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 --> First, we have to find the volume of a sphere formula, which is 4/3 pi r^3. So, the Volume is 4/3 pi x 64 = 268.1m^3 Now we use D=m/v the mass is unknown, but we know D is 3000 and V is 268.1 after multiplying 3000 by 268.1, we get 804,300, which is the mass that we were intially solving for. confidence assessment: 3
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12:33:51 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 --> After multiplying through by pi, my answer was the same. I understand this. self critique assessment: 3
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15:00:48 05-14-2007 15:00:48 `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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NOTES -------> I honestly do not know how to do this. I am all ears on how...
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15:00:56 `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 --> confidence assessment:
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15:04:55 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 --> Okay, first one must find the total grams for each material by multiplying the volume by the density. Then, add them together. This number is the total mass that will be divided by the total volume. The volume should be found by merely combining the volumes from material one and two. Divide the total mass (from step one) by the total volume (in step two), and this gives the total density of the two materials. After reading the explanation this makes sense, and I feel like I can apply it again. self critique assessment: 2
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15:11:30 `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 --> First, one must find the mass of each material by multiplying the density of the material by the given volume. For the first material, the mass is 56,700 kg. For the second, it is 24,000 kg. Then, we find the combined volume which is given twice in the equation--first by the dimensions of the box and second by adding the volumes specifically given for each material. We add the masses of both materials (56,700 +24,000) and get 80,700. Then, we divide this number by the total volume of 30m^2, yielding the total density of 2,690 kg/m^3. confidence assessment: 2
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15:12:36 We find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..
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RESPONSE --> I gave a more specific answer of 2,690, which is still correct. I understand it this time! Yay! self critique assessment: 3
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15:17:29 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
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RESPONSE --> First, one must find the volume by multiplying 1,700,000x1,700,000x.015=4.335x10^10 Then, since we know the final density and the volume, we merely multiply the final density by the volume to give us the mass: 860 kg/m^3 x 4.335 x 10^10 = 3.7281 x 10^13 confidence assessment: 2
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15:19:50 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 see that I multiplied 1,700,000 too many times, messing the equation up from the beginning. I didn't realize (although the squared on the m should have been THE clue) that the figure above already represented the area. The rest of the equation I understand...the numbers were merely messed up from me trying to find the volume and multiplying 1,700,000 too many times. self critique assessment: 2
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15:21:05 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> The surface area of a cylinder can be found by using the formula (2 pi rh) + (2 pi r^2), where the first part of the equation represents the curved sides and the latter part represents the top and bottom of the cylinder. confidence assessment: 3
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15:21:14 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 --> self critique assessment: 3
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15:21:38 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The formula is SA=4pi(r)^2 confidence assessment: 3
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15:21:48 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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RESPONSE --> self critique assessment: 3
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15:22:41 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density involves how much mass there is in a certain amount of space, measured by the term volume. Thus, the formula for density incorporates these ideas with the mass being divided by volume. confidence assessment: 3
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15:22:53 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 understand. self critique assessment: 3
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15:24:10 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> One can divide the mass by the density D=m/v Dv=M V=m/D confidence assessment: 3
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15:24:26 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> I understand how this works. self critique assessment: 3
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15:25:25 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> Like before, I have taken notes thoroughly and have drawn diagrams that supplement my notes. I have documented every example and have explained in writing for myself the areas of the problems with which I had trouble, so that hopefully I will be able to reference this in the future and will not make the same mistakes again. confidence assessment: 3
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