course phy 201
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14:31:51 `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 --> Area of a rectangle is length X width 4 X 3 = 12 The area is 12 meters
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14:32:59 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 --> Remember to include the units and to times them also
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14:35:34 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> Area = 1/2 (hb) = 1/2(4)(3) = 6 meters squared
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14:36:05 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
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14:37:20 `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 --> Area = bh = (5)(2) = 10.0 meters squared
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14:37:44 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
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14:38:55 `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 --> Area = 1/2(b)(h) = 1/2(5)(2) = 5.0 cm squared
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14:39:11 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
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14:46:42 `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 --> A trapezoid can be the same as a parallelogram therefore, Area = bh = (4)(5) = 20.0 km squared
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14:50:21 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 --> right answer - wrong way to look at it - need to look at is as a rectangle not a parallelogram
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14:54:21 `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 --> Looking at a trapezoid as a rectangle, Area - b(avg. of altitudes) Avg of Altitudes= 3(8)/2 = 12 Area = (4)(12) = 48 cm squared
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14:55:36 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 --> the average of altitudes were wrong i is 3 + 8 / 2 NOT 3(8)/2 if I had gotten the avg. altitude right I would of gotten the right answer.
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15:00:39 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> Area of a circle = (pi)r^2 = (pi)3^2 = 28.27 cm^2
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15:02:40 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, you should leave pi in the answer instead of completely multipying it out because the decimal points are approximations.
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15:03:45 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C = 2(pi)r = 2 (pi) (3) = 6pi cm
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15:03:59 The circumference of this circle is C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.
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RESPONSE --> ok
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15:07:02 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> Area = (pi) d^2 / 4 = (pi)144/4 = 36 pi m^2 OR an approximation would be = 113 m^2
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15:07:26 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
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15:10:52 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> C = 14 pi m = 2 (pi) r = 2 (pi) 7 r = 7 Area = pi r^2 = 49 pi m^2
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15:11:04 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 -->
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15:15:27 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> Area = 78 Area = pi r^2 = 78/ pi = 25 =25 square root = 5 m r = approx. 5 m
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15:16:03 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
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15:16:18 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> Length times width
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15:16:32 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
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15:18:43 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> As half of a rectangle = 1/2 (bh)
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15:18:57 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
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15:19:27 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Area = base times hieght
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15:19:39 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> ok
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15:20:44 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> The Area of a trapezoid is the same as a rectangle Area = Hieght times Base
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15:21:09 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> ok - average altitude (hieght) times the width
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15:22:43 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> = pi (r)^2 OR = pi (d)^2 / 4
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15:22:54 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok
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15:24:04 `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 C = 2 pi r To avoid confusing this with the Area of a circle, is looking at the units for the Area, the units are squared for C the units are not squared
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15:24:15 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
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15:26:44 `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 written down all the formulas used and also have written the reasons why they are used and how to visualize the areas that we are calculating
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15:26:57 This ends the first assignment.
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RESPONSE --> ok
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