course Mth 271
003. Misc: Surface Area, Pythagorean Theorem, Density Question: `q001. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
Your solution:
SA = 2ab + 2bc + 2 ac
SA = 2(3)(4) + 2(4)(6) + 2(6)(3)
SA = 108
Confidence Assessment:
OK
Self-critique Rating:
To find the answer, I used the equation for surface area of a rectangular prism
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Question: `q002. What is the surface area of the curved sides of a cylinder whose radius is 5 meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
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Your solution:
C = 2pi r
C = 2pi(5)
C = 31.42
A = C(h)
A = (31.42)(12)
A = 376.99
Confidence Assessment:
OK
Self-critique Rating:
This equation was hard and I had to use the given solution to guide me. Based on the question, I didn’t understand what I was looking for. I vaguely understand how I got the answer I did.
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Question: `q003. What is surface area of a sphere of diameter three cm?
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Your solution:
SA = 4pi r^2
D = 1/2r
D = ½(3)
SA = 4pi (1.5)^2
SA = 28.27
Confidence Assessment:
OK
Self-critique Rating:
To find the solution, I used the equation for SA for a sphere and used solved using the diameter rather than the radius
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Question: `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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Your solution:
a^2 = b^2 + c^2
a = ?
b = 5
c = 9
a^2 = (5)^2 + (9)^2
a^2 = 25 + 81
a^2 = 106
a = 10.29
Confidence Assessment:
Great
Self-critique Rating:
I used Pythagorean’s theorem to solve this problem.
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Question: `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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Your solution:
a^2 = b^2 + c^2
a = 6
b = 4
c = ?
(6)^2 = (4)^2 + c^2
36 = 16 + c^2
20 = c^2
c = 4.47
Confidence Assessment:
OK
Self-critique Rating:
I used Pythagorean’s theorem to solve this problem.
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Question: `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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Your solution:
V = lwh
V = (4)(7)(12) = 336
D = 700/336 = 2.06
Confidence Assessment:
OK
Self-critique Rating:
I used the given solution to guide me because I didn’t understand how to go about finding the answer.
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Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
Your solution:
D = 3000
V = 4/3pi r^3
V = 4/3pi(4)^3
V = 4/3pi(64)
V = 268.08
M = d(v)
M = (3000)(268.08)
M = 8.0 * 10^5
Confidence Assessment:
OK
Self-critique Rating:
There were several steps involved in this problem, and while I completely understand how to go about this problem, I used the given solution to help me.
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Question: `q008. If we build 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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Your solution:
M1 = 6*4 = 24
M2 = 10*2 = 20
Mt = 44
Ad = 44/(6 + 10)
Ad = 44/16 = 2.75
Confidence Assessment:
Alright
Self-critique Rating:
I used the given solution to help with the average density equation.
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Question: `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?
Your solution:
Ad = Mt/Vt
Ms = 27 * 2100 = 56700
Mc = 3 * 8000 = 24000
Ad = (56700 + 24000) / (27 + 3)
Ad = 2690
Confidence Assessment:
OK
Self-critique Rating:
I struggled somewhat with this problem.
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Question: `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?
Your solution:
V = ah
V = (1700000)(.015) = 25500
M = dv
M = (860)(25500)
M = 21930000
Confidence Assessment:
Good
Self-critique Rating:
This problem was relatively simple. I used the area and depth given to find the volume. From there, I used the volume multiplied by the density to find the mass.
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Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder?
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Your solution:
The surface area of a cylinder is equal to 2pi r^2 + 2 pi r h
Confidence Assessment:
Great
Self-critique Rating:
I understand the given equation.
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Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere?
Your solution:
4 pi r^2
Confidence Assessment:
Great
Self-critique Rating:
I understand the given equation
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Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'.
Your solution:
Density is mass divided by volume
Confidence Assessment:
Great
Self-critique Rating:
I understand the meaning of density
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Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume?
Your solution:
Use the equation mass = density * volume and solve for volume
Confidence Assessment:
Great
Self-critique Rating:
I understand how to find volume
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Question: `q015. Part 1 Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
This exercise has served as a refresher course for various equations.
Do include the given solutions with the documents you submit. There are many good reasons for you to do so.