course Mth 173 assignment #003003. Misc: Surface Area, Pythagorean Theorem, Density
......!!!!!!!!...................................
21:56:46 `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 --> Multiply 3 by 4, 4 by 6, and 6 by 3 to constitute each side. Add each number (18, 12, and 24) twice. So 18+18+12+12+24+24= 108 m^2 confidence assessment: 2.5
.................................................
......!!!!!!!!...................................
21:57:00 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:00:21 `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 --> 12 by 10 (2r) gives us the curved side= 120 m^2 78.5 (r^2*3.14) times 2 plus 120 equals total surface area = 277 m^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:02:25 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 --> I should gotten the circumference, not the diameter. Other than that, I understood what to do. So i just need to remember in getting the curved side you multiply height times circumference. confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:04:57 `q003. What is surface area of a sphere of diameter three cm?
......!!!!!!!!...................................
RESPONSE --> The surface area of a sphere is 4 pi r^2. So we use 1.5 as r, squared that is 2.25, times pi that is 7.065, times 4 that is 28.26. So the answer is 28.3 cm^2. confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:05:54 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 --> I got it right, but once again included pi. Not sure which to do still. I know how to do both though, so an easy fix. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:07:46 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
......!!!!!!!!...................................
RESPONSE --> a^2 + b^2 = c^2 (hypoteneuse). So 25 +81= hypoteneuse squared. Squareroot of 106= approx. 10.3 m. confidence assessment: 2.5
.................................................
......!!!!!!!!...................................
22:08:12 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 --> okay self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:10:14 `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 --> well, 4^2 + b^2 = 6^2. 16 + b^2= 36 36-16= 20 then get square root of 20: The answer is approx. 4.5 m. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:11:03 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:12:40 `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?
......!!!!!!!!...................................
RESPONSE --> first volume, which is 4 by 7 by 12= 336 cm^3 density is d=m/v So, fill it in...700/336= 2.08 g/cm^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:14:03 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 --> ok. My last to answers have been a few decimal pts away from yours. The last one was 0.1 off and this one is .02 off. I'm typing it in my calculator correctly, I checked. So i don't know.... self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:16:32 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
......!!!!!!!!...................................
RESPONSE --> Set up the density formula as 3000= x/ (volume= 4*pi*4^2) 200.96 Then multiply 3000 by 200.96= 602,880 kg confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:19:53 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 --> Oh no! I used the surface area formula for a sphere instead of volume. I just wasn't thinking, careless mistake. I know volume is used in density formula and i know the formula for volume of a sphere is (4/3)*pi*r^3. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:28:40 `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 --> Figure mass through density formula: 4= x/6 x=24 2=x/10 x=20 Multiply 24 by 4 and 20 by 2, then add the two answers (this gives us the total density per mass) Then divide 136 by 6 (the two seperate units of density added) The answer is on average, the density of the combined object will be 22 and 2/3. confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:30:52 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 --> Oh my. I got this answer before but I was thinking for some reason average mass. Which is what I got somehow. I understand how I should have added masses to get 44 and then divided it by average volume. Hence the density formula. self critique assessment: 2
.................................................
......!!!!!!!!...................................
22:38:06 `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 --> First, I found the mass of the sand (2100= x/27) The answer was 56700. Then, I found the mass of the cannonballs (8000= x/3) The answer was 24000. Then, I added the masses together (80700) and the volumes (27 + 3 =30). Then, using density formula again I divided 80700 by 30 and the average density is: 2690 kg/m^3 confidence assessment: 2.5
.................................................
......!!!!!!!!...................................
22:38:20 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:42:24 `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 is attained by multiplying 1700000 by the depth, .015. The answer is 25500. The density formula is set up as 860= x/25500. Multiply 25500 by 860 equals mass=21930000 kg. confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:42:34 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:43:18 `q011. Summary Question 1: How do we find the surface area of a cylinder?
......!!!!!!!!...................................
RESPONSE --> circumference of base times height plus area of base times 2. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:43:43 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:43:58 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
......!!!!!!!!...................................
RESPONSE --> 4 pi r^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:44:03 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:44:50 `q013. Summary Question 3: What is the meaning of the term 'density'.
......!!!!!!!!...................................
RESPONSE --> mass divided by volume. How many particles, how close together atoms are, in a given space. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:45:06 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 --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:46:11 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
......!!!!!!!!...................................
RESPONSE --> Mass divided by density equals volume. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:46:26 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:46:44 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
......!!!!!!!!...................................
RESPONSE --> I wrote down the formulas. confidence assessment: 3
.................................................