Orientation Part 4

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course Mth 152

Task: `q001. You have completed the Introductory QA. Please explain the question-answer-self critique process as you understand it at this point.**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

I understand it as you are supposed to choose a number between 1 and 3 to judge how well you think you did on that problem. 1 being the lowest and 3 being the highest.

#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)

Task: `q002. Go to the page http://vhmthphy.vhcc.edu/ and click on your course. Bookmark this page (i.e., add it to Favorites on your Internet browser).

Describe what you see at this page.

**** Your response (insert your response beginning in the next line):

I notice a few paragraphs and then a table filled with assignments.

#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)

Task: `q003. You might not have many questions at this point, but in general any question you pose in your responses should be preceded and followed by a series of at least three question marks. This is so the instructor can quickly identify your questions, and also so the instructor doesn't inadvertently overlook your question.

If you have a question about anything you see in the Orientation, please mark in in the indicated manner and include it in your response.

Please describe how you should annotate questions in your responses, and why it is important to do so:

You are supposed to type three question marks at the beginning and the end of the question so the instructor can find the question easily.

**** Your response (insert your response beginning in the next line; the next line is blankd and doesn't include the #$... prompt):

#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)

Task: `q004. If at any point of the Orientation you have trouble following instructions, let your instructor know the specifics:

• Submit a copy of every instruction related to your question.

• Describe what you do at each step, and what happens as a result. Include a copy of the Address box of your Internet

browser, as it appears at every step.

If you are going astray in the process, this will give your instructor the information necessary to quickly get you back on

track

You should send this information using the Submit Question Form (this is the best form to use because it helps you structure your question and prompts you to provide important information), or the Submit Work Form. Having submitted the form it's also acceptable to send an additional note using email.

Please describe what to do if you have trouble following any of the instructions:

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

So far I believe I am following the instructions adequately.

#$&* (Note that your response was to go into 'the next line'; your response will therefore be inserted before this line, not after. This is obvious when you're looking at the form, but if you've copied the form into a text editor it might be less obvious. Hence this note.)

Task: `q005. On the webpage for your course, click on Assts or Assignments (in some courses (e.g., Liberal Arts Mathematics I and II) the assignments are simply given by a table and there is no Assts link), then on Due Dates (in courses which use a table for assignments, the Due Dates link is provided a few lines above the table), and describe what you see there. Note due dates are updated by the first day of classes; until this update the Due Dates will be for the preceding semester.

Please respond with a statement detailing your understanding of the Due Dates page:

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

I understand that the Due Dates page is to serve as a reminder of what we are covering every week.

#$&* (your response should have gone on the line above this one)

Task: `q006. Return to Assts, take a quick look down the page, and describe what you see. It is recommended that you bookmark this page in your browser (e.g., add it to Favorites) so you can access it easily and quickly.

There are probably things you don't understand right now about the Assignments page. You will get more specific and detailed information in subsequent Orientation assignments. Briefly describe what you do and do not understand about this page.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

I understand that the table provides all of the assignments we are going to cover. One of the things I do not understand is what the “Open Query” means.

#$&* (your response should have gone on the line above this one)

Task: `q007. Click on Info and locate the syllabus or course of study for your course (Liberal Arts Mathematics students click on Course of Study, near the top of your page). Read it and acknowledge below that you have been able to find the syllabus and know how to locate it: Note that you should re-read the syllabus or course of study on or shortly after the first day of class, to be sure you have read the document that will apply to the term in which you are taking the course.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

#$&* (your response should have gone on the line above this one)

Task: `q008. Return to the http://vhmthphy.vhcc.edu page and once more click on the page for your course. Add this page to your bookmarks (e.g., in Internet Explorer, add to Favorites), and acknowledge in your response that you have done so.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

I have added the page to my Bookmarks.

#$&* (your response should have gone on the line above this one)

Task: `q009. On the http://vhmthphy.vhcc.edu page, click on Blackboard, and sign in using the same username and password you use to access the registration system. When the Blackboard screen comes up you should find a 'course' entitled 'Supervised Study Current Semester'. Click on this 'course' and give a brief description of what you see. You need not read all the information that appears, just take a quick look and give a line or two of description. (It is possible that you will be doing the Orientation prior to the beginning of the term, and if so this Blackboard 'course' might not yet be available. You might also be doing this before your Blackboard enrollment is run. If either is the case, be sure to make a note to yourself to do check this page out on the first actual day of classes. You won't need it before then.).

Note that enrollment in Blackboard and enrollment in your course at VHCC are two separate things. Blackboard is 'unofficial' and always needs to be regarded as such.

Please respond with a statement detailing your understanding of the Blackboard Supervised Study 'course', and the unofficial nature of Blackboard.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

I notice the page opens to the Assignment page with a Welcome paragraph.

I understand that Blackboard Supervised Study is just to display our grades and nothing more like posting assignments.

#$&* (your response should have gone on the line above this one)

Task: `q010. We might use various features of this Blackboard page, but the course is not delivered using Blackboard. Blackboard has a number of very good features, but it is of necessity a 'closed' system and not sufficiently flexible or efficient to be the primary vehicle. The one feature we are sure to be using is the Grade Center, where you can check your grades on tests and my comments. Locate the tab for the Grade Center (it might be listed under 'My Grades') and click on it. Describe what you see.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

I see a table of Item Name, Total, and Weighted Total on the Y axis and Item Name, Details, Due Date, Last Submitted, Edited, or Graded, Grade, Points Possible, and Comments on the X axis.

#$&* (your response should have gone on the line above this one)

Task: `q011. You need not look at any more of the Blackboard page, but you are welcome to do so if you choose, and you may ask any questions you wish in your response. If you have no questions, or do not feel the need to further explore this page, just respond with 'ok' or something of that nature.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

ok

#$&* (your response should have gone on the line above this one)

Now highlight and copy your document, paste your copy into the box below, and click on Submit Form. It is suggested that you save a copy of your document as a backup.

"

&#Your work looks good. Let me know if you have any questions. &#

course Mth 173

1/15/11 12:28am

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

003. Misc: Surface Area, Pythagorean Theorem, Density

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Question: `q001. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?

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Your solution:

The surface area of the rectangular solid is 108m^2.

confidence rating #$&*: 3

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Given Solution:

`aA 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `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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Your solution:

The surface area of the sides of the cylinder is 376.99m^2. The total surface area of the cylinder is 534.1m^2.

confidence rating #$&*: 3

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Given Solution:

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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Self-critique (if necessary):

I only used the actual value of the solution in my answer instead of the process to get my answer. I understand the arithmetic used to get the answer because I used the same arithmetic and wrote the problem in a notebook.

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Self-critique Rating: 2

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Question: `q003. What is surface area of a sphere of diameter three cm?

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Your solution:

The surface area of the sphere is 28.27 cm^2. 4*pi*(1.5)^2= 4*pi*2.25 = 9*pi

confidence rating #$&*: 3

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Given Solution:

`aThe 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.

NOTE TO STUDENT:

While your work on most problems has been good, you left this problem blank and didn't self-critique.

You should self-critique here.

• For example you should acknowledge having made note of the formula for the surface area of the sphere, which I expect you didn't know before.

I expect from your previous answers that you are very capable of applying the formula once you have it, and based on this history you probably wouldn't need to self-critique that aspect of the process.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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:

The hypotenuse is 10.3 meters^2

confidence rating #$&*: 3

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Given Solution:

`aThe 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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:

The length of the other leg is 4.47m, or sqrt(20)

confidence rating #$&*: 3

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Given Solution:

`aIf 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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:

The density of the rectangular solid is 2.08 g/cm^3

confidence rating #$&*: 3

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Given Solution:

`aThe 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 (for example 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).

NOTE TO STUDENT: (in this note the instructor attempts to clarify the idea of 'demonstrating what you do and do not understand about the statement of the problem' and 'giving a phrase-by-phrase analysis of the given solution')

You did not respond to the question and did not self-critique.

You would be expected to address the question, stating what you do and do not understand.

• For example you should understand what a rectangular solid with dimensions 4 cm by 7 cm by 12 cm is, and how to find its volume and surface area. You might not know what to do with this information (for example you might well not understand that it's the volume and not the surface area that's related to density), but from previous work you should understand this much, and should at least mention something along the lines of 'well, I do know that I can find the volume and/or surface area of that solid' in a partial solution.

• The word 'density' is clearly very important. Even if you don't know what density is, you could note from the statement of the problem that its units here are said to be 'grams per cubic centimeter'.

Having noted these things, you will be much better prepared to understand the information in the given solution.

Then you need to address the information in the given solution. A 'phrase-by-phrase' analysis is generally very beneficial:

• I expect you understand the first statement from previous knowledge (you should have this understanding from prerequisite courses, and if not you encountered it in the preceding 'volumes' exercise): 'The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.' It would of course be appropriate to ask a question here if necessary.

• It is likely that, as is the case with many students, the concept of density is not that familiar to you. However if this wasn't addressed specifically in prerequisite courses, those courses would be expected to prepare you to understand this concept. The statement 'Its density in grams per cm^3 is the number of grams in each cm^3.' serves as a definition of density. In your self-critique you should have addressed what what this phrase means to you, and what you do or do not understand about it

• The next phrase is 'We find this quantity by dividing the number of grams by the number of cm^3.' You would be expected to understand that this phrase is related to the preceding, and as best you can to address the connection. At this point many students would need to ask a question, and it would be perfectly appropriate to do so (or to have done so regarding previous statements).

• The subsequent phrase 'density = 700 grams / (336 cm^3) = 2.06 grams / cm^3' is an illustration of the ideas and definitions in the preceding statements. A reasonable self-critique would demonstrate your attempt to understand this statement and its connection to the preceding. Once again questions would also be appropriate and welcome.

• The above addresses sufficient information to solve the problem. If you get to this point, you're probably doing OK and you wouldn't necessarily be expected to address the rest of the given solution, which expands on the finer details of the problem and provides additional information. The basic prerequisite courses should have prepared you to understand the information, but students entering Liberal Arts Mathematics, College Algebra and even Precalculus or Applied Calculus (or Physics 121-122) courses probably don't need to address anything beyond the basic solution at this point. Though Precalculus and Applied Calculus students could benefit from doing so, and if time permits would certainly be encouraged to do so, time is also a factor and it would be understandable if these students chose to move on.

• Students entering the Mth 173-4 sequence or the Phy 201-202 or 231-232 sequence would be expected to either completely understand all the details of the given solution, or address them in your self-critique.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The mass of the sphere is 804240 kg.

confidence rating #$&*: 3

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Given Solution:

`aA 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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:

6cm^3*4g/cm^3= 24g, 10cm^3*2g/cm^3= 20g, 24+20= 44g. The combined volume is 6cm^3+10cm^3= 16cm^3. The average density is 44g/16cm^3= 2.75g/cm^3.

confidence rating #$&*: 3

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Given Solution:

`aThe 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The total mass is found by finding the mass of the sand (2100kg/cm^3*27cm^3 = 56700kg) and finding the mass of the cannon balls (8000kg/cm^3*3cm^3 = 24,000kg) and adding the two massed together (56700kg + 24,000kg = 80,700kg). Then, the total mass is divided by the total volume to get the average density, 80,700kg/30cm^3 = 2,690kg/cm^3.

confidence rating #$&*: 3

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Given Solution:

`aWe 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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?

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Your solution:

The volume of the oil slick (a cylinder) is 25,500m^3. The density of the oil slick is 860kg/m^3, so the mass of the oil slick is 21,930,000 kg.

confidence rating #$&*: 3

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Given Solution:

`aThe 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 * 25,500 m^3 = 21 930 000 kg.

This result should be rounded according to the number of significant figures in the given information.

STUDENT QUESTION

I didn’t round to the most significant figure. ???? How important is this?

INSTRUCTOR RESPONSE

It will be important.

This document is preliminary; the issue of significant figures will be addressed more specifically as we move into the course.

Right now I just want you to be aware of the general idea.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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 its radius multiplied by pi then doubled, then added to the product of its radius multiplied by its height and multiplied by pi and doubled.

confidence rating #$&*:

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Given Solution:

`aThe 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere?

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Your solution:

The surface area of a sphere is found with the formula 4*pi*r^2

confidence rating #$&*: 3

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Given Solution:

`aThe surface area of a sphere is

A = 4 pi r^2,

where r is the radius of the sphere.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'.

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Your solution:

Density is mass with respect to volume. D=m/v

confidence rating #$&*: OK

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Given Solution:

`aThe 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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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume?

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Your solution:

If the average density and mass is known, volume is found by dividing the mass by the density.

confidence rating #$&*: 3

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Given Solution:

Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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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.

I have written all of the formulas in an assigned notebook for this class, to be organized.

"

&#This looks good. Let me know if you have any questions. &#