course Phy202 {qúwassignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 08-20-2008
......!!!!!!!!...................................
15:57:29 `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?
......!!!!!!!!...................................
RESPONSE --> The top and the bottom of the rectangular solid has the measurements of 3 meters by 4 meters. 2(3m * 4m) = 24 m^2 The front and back has measurements 4m by 6 m. 2(4m*6m) = 48 m^2 The sides have measurements of 3m by 6m. 2(3m*6m) = 36 m^2 So, 24m^2 + 48m^2 + 36 m^2 = 108 m^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:57:48 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:04:23 `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?
......!!!!!!!!...................................
RESPONSE --> The formula for the surface area of the curved side is: SA = 2*Pi*r*h SA = 2*3.14*5m*12m SA = 6.28*60m^2 SA = 376.80 m^2 confidence assessment:
.................................................
ԧ~Qz assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 08-20-2008 Sκyw assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 08-20-2008 ؛ƭk assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 08-20-2008
......!!!!!!!!...................................
16:08:26 `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?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
16:08:33 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
16:17:33 `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
......!!!!!!!!...................................
RESPONSE --> The surface area of the curved side is: SA = 2*Pi*r*h SA = 2*3.14*5m*12m SA = 6.28*60m^2 SA = 376.80 m^2 (of curved side) The surface area of the top and bottom (which is a circle) is each: SA = Pi*r^2 SA = 3.14*(5m)^2 SA = 3.14 (25m^2) SA = 78.5 m^2 For the top and the bottom, we have to multiply by 2: 2(78.5m^2) = 157m^2 So, we add the SA of the curved side to the top and bottom to get the total SA: 376.80m^2 + 157m^2 = 533.8 m^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:18:58 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.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:22:55 `q003. What is surface area of a sphere of diameter three cm?
......!!!!!!!!...................................
RESPONSE --> Since the diameter is 3 cm, then the radius is 1.5 m. SA of a sphere is 4*Pi*r^2 SA = (4)(3.14)(1.5m)^2 SA = (12.56)(2.25m^2) SA = 28.26 m^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:23:20 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.
......!!!!!!!!...................................
RESPONSE --> Understood. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:26:57 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
......!!!!!!!!...................................
RESPONSE --> We can use the Pythagorean Theorem: a^2 + b^2 = c^2 (5m)^2 + (9m)^2 = c^2 where c is the hyp. 25m^2 + 81m^2 = c^2 106m^2 = c^2 we now take the square root 10.3 m = c confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:27:17 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.
......!!!!!!!!...................................
RESPONSE --> Understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:31:12 `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?
......!!!!!!!!...................................
RESPONSE --> We can still use the Pythagorean Theorem and just solve it for b. a^2 + b^2 = c^2 where 6m = c and 4m=a (4m)^2 + b^2 = (6m)^2 16m^2 + b^2 = 36m^2 sub. 16m^2 from both sides b^2 = 20m^2 take the sqrt. of 20m^2 b = 4.47m (approx.) confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:31:30 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.
......!!!!!!!!...................................
RESPONSE --> Understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:35:03 `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?
......!!!!!!!!...................................
RESPONSE --> Density = Mass/Volume D = 700g/(4cm*7cm*12cm) D = 700g/336cm^3 D = 2.08 g/cm^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:35:33 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).
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:48:02 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
......!!!!!!!!...................................
RESPONSE --> Volume = Mass/Density It follows that, Mass = Volume * Density The volume of a sphere is: V = 4/3*Pi*r^3 V = (4/3)(22/7)(4m)^3 V = (4/3)( 22/7)(64m^3) V = (256/3m^3)(22/7) V = 5632/21 m^3 V = 268.19 m^3 approx. Since, Mass = V*Density Mass = 268.19 m^3 * 3,000kg/m^3 Mass = 804,570 kg confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:49:10 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.
......!!!!!!!!...................................
RESPONSE --> Understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:59:52 `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?
......!!!!!!!!...................................
RESPONSE --> Density = Mass/Volume The first material: 4g/cm^3 = M/6cm^3 24g = Mass The second material: Density = Mass/Volume 2g/cm^3 = Mass/10cm^3 20g = Mass Density of first material: Density = Mass/Volume Density = 24g/6cm^3 Density = 4g/cm^3 Density of second material: Density = Mass/Volume Density = 20g/10cm^3 Density = 2g/cm^3 The average density is: (4g/cm^3 + 2g/cm^3)/2 3g/cm^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:01:21 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.
......!!!!!!!!...................................
RESPONSE --> I did not consider the total mass and the total volume. self critique assessment: 1
.................................................
......!!!!!!!!...................................
17:19:24 `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?
......!!!!!!!!...................................
RESPONSE --> Density = M/V Average Density = Total Mass/Total Volume We know the Volume of the Sand is 27m^3 and the Density of the sand is 2,100 kg/m^3 Mass = Density*Volume Mass = (2100kg/m^3)(27m^3) Mass = 56,700kg We know the Volume of the Cannon Balls is 3m^3 and the Density of the Cannon Balls is 8,000 kg/m^3 Mass = Density *Volume Mass = (8,000 kg/m^3)(3m^3) Mass = 24,000kg Average Density = Total Mass/Total Volume Average Density = (56,700kg + 24,000kg)/(27m^3 + 3m^3) Average Density = 80,700kg/30m^3 Average Density = 2,690kg/m^3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:19:49 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..
......!!!!!!!!...................................
RESPONSE --> Understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:24:08 `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?
......!!!!!!!!...................................
RESPONSE --> The volume of oil is: 1,700,000m^2 * .015m = 25,500m^3 Mass = Density * Volume Mass = (860kg/m^3)(25,500m^3) Mass = 21,930000 kg confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:25:12 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.
......!!!!!!!!...................................
RESPONSE --> I think we should have multiplied by 25,500 instead of 24,400 like is shown in the answer. Please let me know. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:30:01 `q011. Summary Question 1: How do we find the surface area of a cylinder?
......!!!!!!!!...................................
RESPONSE --> We must find the SA of the curved side, the top, and the bottom of the cylinder. The top and the bottom are both circles. To find the area of a circle, we use? A = Pi*r^2 To find the area of the curved side, we use? SA = 28Pi*r*h Then we add the SA of the top + the SA of the bottom + the SA of the curved side. We can keep Pi in the answer for an exact answer of we can approximate the answer by substituting 3.14 for Pi. confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:30:39 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.
......!!!!!!!!...................................
RESPONSE --> Understood self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:31:37 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
......!!!!!!!!...................................
RESPONSE --> SA = 48Pi*r^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:32:02 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> I should have explained r in the answer I gave. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:36:01 `q013. Summary Question 3: What is the meaning of the term 'density'.
......!!!!!!!!...................................
RESPONSE --> The density is the heaviness of objects with a constant volume. For example, an oil spill in the water....The oil floats on top of the water, so the oil is less dense than water. Density can also refer to how tightly packed an object is. confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:36:39 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'
......!!!!!!!!...................................
RESPONSE --> I should have been more specific. self critique assessment: 1
.................................................
......!!!!!!!!...................................
17:38:39 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
......!!!!!!!!...................................
RESPONSE --> We take the density and divide it by the mass to get the volume. We can use the formula V = D/M and manipulate the formula to find any of the three: Volum, Density, or Mass confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:39:06 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
......!!!!!!!!...................................
RESPONSE --> Understood. self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:41:29 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I need to remember the units and make sure the calculations come out with the correct units. This usually helps with how to find volume...because it is in units^3. The mass is in a weight such as kg. Density is a weight/volume. confidence assessment: 2
.................................................