course Phy 121 I am SO sorry. I know this is late, but I had time to look over my access page and my vhmthphy file and compare everything. I noticed that this q_a for assignment 3 was not posted. Does that mean I did not send it or could it be that it just didn't get to my access page? I always check things off when I send them, and this assignment was checked off. If I didn't turn it in, is it too late? 1???Q????????????Student Name:
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23:27:29 `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 --> Surface Area = 2(lenght)(width) + 2(length)(height) + 2(width)(height) S.A. = 2(3*4) + 2(4*6) + 2(3*6) = 108 m^2
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23:27:55 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
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RESPONSE --> I understand this problem.
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23:28:54 `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 --> For a closed cylinder: Area of base = pi r^2 = pi * (5 m)^2 = 25pi m^2 2 bases so, 2(25pi m^2) = 50pi m^2 Area of sides = 2pi (radius x height) = 2pi(5m*12m) = 120pi m^2 Total Surface Area = 50pi m^2 + 120pi m^2 = 170m^2
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23:34:45 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 --> Took me a while, but I think I understand this problem.
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23:36:29 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> Surface area for sphere= 4pi r^2 = 4pi (3/2 ^2) = 4pi (1.5^2) = 4pi x 2.25 = 9pi cm^2
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23:36:37 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 --> Ok
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23:39:38 `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 5m^2 + 9m^2 = c^2 25m^2 + 81m ^2 = c^2 106m^2 = c^2 TAKE SQUARE ROOT 10.30 m = c
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23:39:47 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
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23:41:44 `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 --> a^2 + b^2 = c^2 c^2 - a^2 = b^2 6m^2 - 4m^2 = b^2 36m^2 - 16m^2 = b^2 20m^2 = b^2 TAKE SQUARE ROOT 4.47m = b
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23:41:50 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 --> Ok
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23:43:13 `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 D = 700g / (4cm x 7cm x 12cm) = 2.08 g/cm^3
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23:43:24 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 --> Ok
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23:47:08 `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 --> d = m/v m = v*d mass = (4/3pi (4m^3))(3000kg/m^3) = (4/3pi x 64m^3)(3000kg/m^3) = (85.33pi m^3)(3000kg/m^3) = 256000pi kg
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23:47:17 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 --> Ok
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23:49:18 `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 --> First we must find the total mass and total volume. Mass = ( 4g/cm^3 x 6cm^3) + (2g/cm^3 x 10cm^3) = 44grams Volume = (6cm^3 + 10cm^3) = 16cm^3 Density = 44g / 16cm^3 = 2.75g/cm^3
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23:52:10 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 --> Ok
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23:52:27 `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 we must find total mass and total volume. Mass = (27m^3 x 2100kg/m^3) + (3m^3 x 8,000kg/m^3) = 56700kg + 24000kg = 80700kg Volume = 27m^3 + 3m^3 = 30m^3 Density = m/v = 80700kg / 30m^3 = 2690 kg/m^3
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23:56:27 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
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23:56:40 `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 --> Volume = Area x altitude = (1700000m^2)(.015m) = 25500 m^3 Mass = density x volume = (860kg/m^3)(25500m^3) = 21930000 kg
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23:56:45 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 --> ok. Where did your 24400 m^3 come from?
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23:56:58 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> Surface area of a cylinder = Area of sides + area of bases = 2pi(rxh) + 2pi r^2
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23:57:11 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
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23:57:14 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> Surface area of sphere = 4pi r^2
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23:57:18 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 --> OK
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23:57:41 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is a measure of the mass in proportion to the volume of an object. density = mass/volume
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23:57:46 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. NOTE: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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23:58:15 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> density = mass/volume Volume = mass / density
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23:58:20 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> Ok
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23:58:40 `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 added these formulas and ideas to my formula sheet.
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