Explain which dimensions you will must take and extra info that you may need in order to calculation the thickness of a soda can. You might assume that the have the right to is make of aluminum. Because of the risk of injury, cutting the can and directly measure its thickness is not permitted.

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## IM Commentary

This is a sports of \"\"How thick is a soda can? sports I\"\" which allows students to work-related independently and also think around how they deserve to determine exactly how thick a soda have the right to is. The teacher should explain plainly that the goal of this job is come come up through an \"\"indirect\"\" way of assessing how thick the deserve to is, that is straight measuring that thickness is not allowed. Students must be provided with cans to aid in this process. After permitting the students to think about this because that a while, it would certainly be good to have a discussion about what tactics students have developed and what measurements they think of taking. The many relevant dimensions for this difficulty are surface ar area, weight, and density so the teacher might wish to discuss these if they are not brought up by the students.

If student have access to ice measures and relatively finely calibrated scales, then these deserve to be used and would enable for a discussion around precision in this measurements. As far as the density of aluminum goes, the instructor might wish to provide this information to college student ($2.70$ grams per cubic centimeter). So the key pieces of info which the students may identify as being essential to calculation the thickness that the have the right to are:

the surface ar area of the can, the load of the can, the density of the can.

Assuming that the thickness of the can is uniform, the students would certainly then usage the formula: $$\\rm density \\approx \\frac\\rm surface\\,\\,\\, \\rm area \\times \\rm thickness\\rm weight$$ to find the approximate thickness. The factor why the $\\approx$ is ideal in this vault formula, instead of $=$, is the we perform not know that the thickness the the have the right to is uniform and also the \"\"edges\"\" that the deserve to will likewise influence this relationship. Moreover, soda cans space not in reality cylindrical as they room slightly tapered both at the top and bottom.

A second technique for estimating the thickness the the have the right to uses Archimedes\" principle, gift in \"\"Archimedes and also the King\"s crown,\"\" and also this walk not call for the density. Archimedes\" technique calculates the volume of aluminum in the deserve to by submerging the in water in a calibrated beaker and also measuring just how much water is displaced. This is perhaps more appropriate and also feasible in a chemistry class but the ideas can be disputed even if the is not possible to execute the measurements in math class. Most importantly, Archimedes\" technique avoids learning the thickness of aluminum and additionally the load of the can. So making use of this method, the students would certainly then usage the formula $$\\rm Volume \\approx \\rm surface\\,\\,\\, \\rm area \\times \\rm thickness$$ The $\\approx$ sign is suitable both due to the fact that the surface area and volume that the have the right to are both estimates and also because the wherein the top and also bottom that the have the right to are joined v the cylindrical shaped piece, closer modeling would certainly be necessary to get specific formula.

## Solutions

Solution:1 Use thickness of aluminum

A picture of the soda can is offered here. This deserve to be mutual with the studentsif they do not have access to soda can be ~ or come the appropriate measuring tools that castle will need to estimate that is thickness. The only part of an aluminum can which offers us a visual sense for just how thick the can could be is the opened from which we drink the soda. Feeling and observing this,it is fairly sharp and clearly very thin, certainly too slim to it is in measured through a common tape measure. The thickness of the deserve to will, however, be proportional to just how much aluminum is in the can and also this, in turn, can be recorded by weighing the can. Inaddition to the load of the can, the student will need to recognize the thickness of aluminum (that is, just how much does a provided amount the aluminum weigh) and also they willneed to almost right the surface ar area that the can.

The surface of the have the right to is consisted of of the top and bottom in addition to thecylindrical component of the can. In practice, the top and also bottom are not flatbut room well approximated through circles. Follow to the offered information,the radius of this circles is around $1 \\frac316$ inches. Therefore the area isabout $\\pi \\times \\left(1 \\frac316\\right)^2$ square inch or about $4.4$ squareinches. If the cylindrical part of the have the right to were do flat, it would certainly be a rectangle v dimensions $4 \\frac34$ customs by $\\pi \\times 2 \\times1 \\frac316$ inches or about $35.4$ square inches. For this reason the complete surfacearea that the have the right to is about $44.2$ square inches.

If over there are about $15$ grams of aluminum in the soda can and the thickness ofaluminum is about $2.70$ grams every cubic centimeters climate there room about$$\\frac15 \\mbox grams2.70 \\mbox grams per cubic centimeter \\approx 5.6 \\mbox cubic centimeters$$of aluminum in the soda can.

Since the amount of aluminum is offered in cubic centimeters and the area of thesoda can in square inch we need to make a counter in order to estimate the almost right thickness that the can. There are about 2.54centimeters per inch and also so there are around $(2.54)^2 \\approx 6.5$ squarecentimeters per square inch. So$$44.2 \\mbox square inches \\approx 44.2 \\times 6.5 \\mbox square centimeters.$$This is about $287.3$ square centimeters. To discover the almost right thicknessof the deserve to we understand that$$\\rm thickness \\times 287.3 \\mbox cm^2 \\approx 5.6 \\mbox cm^3.$$This means that the approximate thickness the the can is about $0.02$ centimetersor $0.2$ millimeters.

If the teacher desire to connect students in a conversation of precision that measurements and also how results must be recorded, there space three measurementswhich room subject to error and the reported density of aluminum i beg your pardon isonly exact the nearest one hundredth that a gram per cubic centimeter. Themeasurements the the can, if made through a ice measure marked to the nearestsixteenth of one inch, deserve to be assumed come be accurate to in ~ $\\frac132$ ofan inch. This is about $2.5$ percent the the measure of the radius and seventenths of one percent of the measurement for the height. Recognize the approximatearea the the top and bottom of the deserve to requires squaring the radius which willroughly dual the error to $5$ percent. We have actually that $5$ percent of $8.8$square inches is almost fifty percent a square inch. The error for the cylindrical component ofthe have the right to will depend on the error because that the height and also the error for the circumferenceand will be a little an ext than $3$ percentor a little much more than one square inch. So the total feasible error hereis close to one and a half square inches. One appropriate method to document this wouldbe $44 \\pm \\frac32$ square centimeters and also then this error proceeds throughthe final estimation that the thickness the the aluminum can. The canis no perfectly cylindrical and also it has a tab to open up the can: these will contributeto the error as well but the as whole estimate of $0.2$ mm for thickness have to be great provided this thickness is uniform.

Solution:Solution 2 usage Archimedes\" principle

In order come estimate just how thick the soda have the right to is, we can estimate the surfacearea of the can and also the volume of the aluminum in the can and then usage the fact that the volume the aluminum in the can is around equal to the surface ar area of the deserve to times that thickness. For the surfacearea, the method of the very first solution deserve to be used and we repeat this here.

The surface ar of the can is comprised of the top and bottom together with thecylindrical component of the can. In practice, the top and also bottom space not flatbut room well approximated by circles. According to the provided information,the radius of these circles is about $1 \\frac316$ inches. For this reason the area isabout $\\pi \\times \\left(1 \\frac316\\right)^2$ square customs or around $4.4$ squareinches. If the cylindrical component of the have the right to were make flat, it would be a rectangle v dimensions $4 \\frac34$ customs by $\\pi \\times 2 \\times1 \\frac316$ inches or around $35.4$ square inches. So the full surfacearea of the deserve to is about $44.2$ square inches. Over there are about 2.54centimeters per inch and also so there are about $(2.54)^2 \\approx 6.5$ squarecentimeters every square inch. So$$44.2 \\mbox square inches \\approx 44.2 \\times 6.5 \\mbox square centimeters.$$This is around $287.3$ square centimeters.

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The volume that the have the right to is not given and also so this needs to it is in assessed by part indirect means. Making use of a method dating back to Archimedes, if us submerge the have the right to in water, then the lot of water displaced through the deserve to will it is in equalto that volume. One way to do this would certainly be to crush the deserve to (although some care is needed here to make sure that in for this reason doing we perform not trap too much air within the can). Another technique would be to very closely fill the have the right to to the optimal with water and also then the filled can will sink. The lot of water displaced by the aluminum in the have the right to should it is in close to $5.6$ cubic centimeters together wasfound in the very first solution. Speculative values might differ, however, so suppose we find that the aluminum in the soda can displaces $x$ cubic centimeters of water. Then making use of the equation$$\\rm Volume \\approx \\rm surface\\,\\,\\, \\rm area \\times \\rm thickness$$we have the right to plug in $x$ for the volume and $287.3 \\rm cm^2$ for the surfacearea and also we find$$\\rm thickness \\approx \\fracx287.3 \\rm\\, cm^2.$$

Notice the this technique is quicker than the an initial and requires less information. Top top the various other hand, the does need some rap equipment.