course Phy 201 09/23/09 at 12 pm q_a_areas_units_asst3course Phy 201
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14:22:45 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> I believe you just multiply to get the surface area so 3*4*6=72m confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:27:18 02-09-2009 14:27:18 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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NOTES -------> I messed up on this question need to remember that a rectangular solid has six faces(bottom, front, back top, left/right side.
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14:34:18 `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 --> Im not really sure how to do this problem. I think that to get the surface area of the sides you take 5m * 12 m=60m^2 Then you multiply it by 2 in order to get the opposite side so it would be 120m^2 for the sufrace area of the curved sides. And I dont have a clue how to get the total surface area if the cylinder is closed. Maybe multiply it by 3 for three sides?? confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:40:39 02-09-2009 14:40:39 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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NOTES -------> I was thinking of a cylinder like a cone for some reason.to get the total area you add the area of the sides with 2* the area of the base.
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14:46:05 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> The surface area of a sphere is Area= 4pi r^2. So since the diameter is three cm then the radius would be 1/2 the diameter so 1.5. Then A=4*(3.14)*1.5^2=28.26cm^2 confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:49:30 02-09-2009 14:49:30 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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NOTES -------> I had the equation right but I messed up on the pi. You dont actually let pi = the 3.14 in these problems you just add the pi to the end. Don't multiply out the pie in numbers.
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14:56:56 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> I think you use the pythagorean theorem in this problem. adj^2 + opp^2 =hyp^2 so 5m^2 +4m^2=hyp^2 Which gives 25m^2+16m^2=41m^2 confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:01:46 02-09-2009 15:01:46 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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NOTES -------> I thought the problem had 5 meters and 4 meters instead of 9 it was a silly mistake and I didnt take the sqrts
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15:09:43 `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 --> Again use the pythagorean theorem adj^2+opp^2=hyp^2 So 2m+b=3m=b=3m-2m=1m confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:13:18 02-09-2009 15:13:18 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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NOTES -------> I messed up on the math again. Took the sqrt of the sides before I squared them. Thought that they were already squared.
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15:24:58 `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 --> First you need to determine what the volume of the rectangular solid is so start by multiplying the demensions 4cm*7cm=28cm^2 and then 4cm*12cm=48cm^2 then 7cm*12cm=84cm^2 then add these together to get 160cm^2. After that you get the density= mass/volume. so the mass is 700g/160cm. I think so the density is going to be 4.375. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:30:29 02-09-2009 15:30:29 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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NOTES -------> I messed up again and did the surface area of the rectangular solid instead of multiplying the deminsions together to get cm^3 I needed to get the volume of the rectangular solid
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15:43:24 `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 you need to find the volume of the sphere and valume of a sphere is v=4/3pir^3 so v=4/3 *3.14* 4m^3=4/3*4.13*64=352.43 So then using density= mass/volume you get 3000kg=mass/352.43=3000kg*352.43=mass so mass = 1057290kg/cubic meter. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:53:15 02-09-2009 15:53:15 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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NOTES -------> Well i got the equations right for this problem but i messed up on the calculations completly
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16:02:38 `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 --> Density = mass/volume so 4g/cm^3 = mass/6cm^3=24cm^3 And then you have 2grams/cm^3=mass/10cm^3=2*10=20cm^3. Then you need the average so you add them both together and divide by 2 ao 24+20=44g/cm^3 then divide by 2 gives you 22grams/cm^3 as the average density. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:07:12 02-09-2009 16:07:12 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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NOTES -------> I just divided by 2 instead of taking the total mass and dividing by the total volume.
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16:17:39 `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 --> The average density = total mass/total volume. So the total mass is 2100kg+8000kg =10100kg. Then you get the volume by multiplying the 2*3*5=30m^3. So the average density is 10100kg/30m^3=336.666kg/m^3 confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:24:29 02-09-2009 16:24:29 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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NOTES -------> I messed up on the calculations I did not multiply by the amount of sand or the cubic meters of cannon balls. with the densitys.
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16:37:37 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
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RESPONSE --> The density = mass/volume so 860 kg/cubic meter=mass/1700000*.015=860kg/cubic meter*25500=mass=21,930,000kg/cubic meter confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:44:42 02-09-2009 16:44:42 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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NOTES -------> I think my answer ended up with and extra place holder. I did my calculations wrong again. Think I'm getting a better understanding of these problems. I believe you were right on this one.
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16:48:54 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> Get the surface area of the cylinder is Area=circumference*altitude confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:50:18 02-09-2009 16:50:18 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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NOTES -------> I understand
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16:50:45 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> A=4 pi r^2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:52:34 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. OK. Density is defined as the mass per unit volume: density = m/V confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:54:15 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. ok Density = mass/volume so to get the volume you can multiply the density and the mass. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:55:23 02-09-2009 16:55:23 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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NOTES -------> you need to divide the mass/ density to get the volume not multiply
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16:56:47 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> This was a helpful assignment. I don't think that I completly got any answer right. But I do have a better understanding of the exercise. I took notes on every question and will refer back to them as needed. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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You appear to know what you're doing; you're just making errors in calculation, not in concept. You'll be able to fix this with just a little more attention to the details. "