More problems that combine several ideas. As before, try each part on your own first, and only then open the two boxes below it: the setup (starting point, goal and conversion information) and the computation.
14.1 Tevye the Milkman
In the stories by the writer Sholem Aleichem about Tevye the Milkman (which are the basis of Fiddler on the Roof), Tevye lives in a small village near the town of Anatevka. Life in the Russian Empire was measured with very different units from the ones that we use today. In this exercise we’ll convert some of them.
a. a walk to the butcher
When Tevye visits Leizer-Wolf the butcher, he walks a distance of 3 versts from his home in the village to Anatevka. A verst (верста) was a unit of length used in the Russian Empire, and it equals 500sazhens. A sazhen (сажень) is the distance between the fingertips of the outstretched arms of a (very tall) man, and it equals 2.1336 \text{ m}. If the length of Tevye’s step is 67 \text{ cm}, how many steps does he take from his home to Anatevka? (Round to the nearest whole number.)
Along the way we found that the distance is 3 \cdot 500 \cdot 2.1336 \approx 3200 \text{ m}, a bit more than three kilometers. Tevye takes almost five thousand steps to get to town.
b. tevye’s farm, in square meters
Tevye owns a farm with an area of 0.7 acres. An acre is the area of a square with sides of 70 yards, and a yard is 91 \text{ cm}. What is the area of Tevye’s farm in square meters?
Tipsetup: starting point, goal, conversion
This is an area, so we need to be careful: if 1 \text{ yd} = 0.91 \text{ m}, then a square yard is 1 \text{ yd}^2 = (0.91 \text{ m})^2, and the conversion factor is squared. This is the same idea as with the prefixes, when we converted \text{km}^2 to \text{m}^2 (see areas and volumes).
In the Russian Empire, land was measured in desyatinas (десятина). One desyatina is 3200 square sazhens, where a sazhen is 2.1336 \text{ m}. What is the area of Tevye’s farm in desyatinas?
Tipsetup: starting point, goal, conversion
We start from the area in \text{m}^2 that we found in part b. We need a conversion from \text{m}^2 to square sazhens, and again the factor must be squared.
Tevye’s farm is about a fifth of a desyatina. And indeed, in the stories he isn’t a wealthy landowner!
14.2 a sheet of paper
A sheet of paper looks like a very simple object, but we can learn a lot about it from just a few numbers. In this exercise we’ll find out how thick it is, how big it is, and how much a package of paper weighs.
a. how thick is a sheet?
A typical printer paper has an area density (the mass per unit area) of 80 \text{ gsm}, which stands for 80 grams per square meter. The volume density of the same paper is 800 \text{ kg/m}^3. What is the thickness of one sheet, in micrometers (\mu\text{m})?
Tipsetup: starting point, goal, conversion
Let’s think about a piece of paper with an area of exactly 1 \text{ m}^2. Its mass is 80 \text{ g}, and from the volume density we can find how much volume this mass takes. This volume is a slab with an area of 1 \text{ m}^2, so its height is the thickness of the paper: for a slab of area 1 \text{ m}^2, the number of \text{m}^3 is the same as the number of meters of thickness.
Starting point:80 \text{ g} (of a piece of paper of 1 \text{ m}^2).
Goal: the volume of this paper in \text{m}^3, which gives the thickness in meters, and then in \mu\text{m}.
A sheet of paper is about 0.1 \text{ mm} thick, roughly the thickness of a human hair. As a sanity check, 500 sheets would make a stack of 500 \cdot 0.1 \text{ mm} = 5 \text{ cm}, which is indeed how tall a package of paper is.
b. the area of an A4 sheet
The A series of paper sizes follows a simple rule. An A0 sheet has an area of 2^0 \text{ m}^2 (that is, one square meter), an A1 sheet has an area of 2^{-1} \text{ m}^2 (half a square meter), an A2 sheet has an area of 2^{-2} \text{ m}^2 (a quarter of a square meter), and so on. What is the area of a regular A4 printer sheet, in square centimeters?
Tipsetup: starting point, goal, conversion
The rule says that every step in the series divides the area by 2 (if you cut an A0 sheet in half, you get two A1 sheets). So an An sheet has an area of 2^{-n} \text{ m}^2. See the power rules for negative exponents.
Starting point: the area of an A4 sheet, 2^{-4} \text{ m}^2.
The real A4 sheet has sides of 21 \text{ cm} and 29.7 \text{ cm}, and 21 \cdot 29.7 \approx 624 \text{ cm}^2. Very close!
c. a package of paper
What is the mass of a typical package of printer paper? It is A4 paper with an area density of 80 \text{ gsm}, and a package holds 500 sheets. Give the answer in kilograms.
Half a thousand sheets of paper weigh two and a half kilograms, and that’s why a box of paper feels so heavy.
d. american paper
In the United States, the common paper size is not A4, but “letter”, whose dimensions are 8.5 inches by 11 inches. Given that 1 \text{ inch} = 2.54 \text{ cm}, what is the area of a letter sheet, in square meters?
Tipsetup: starting point, goal, conversion
This is an area, so the conversion factor from inches to centimeters must be squared: 1 \text{ in}^2 = (2.54 \text{ cm})^2.
Starting point: the area of the sheet, 8.5 \text{ in} \times 11 \text{ in} = 93.5 \text{ in}^2.
A letter sheet is about 603 \text{ cm}^2, a bit smaller than the 625 \text{ cm}^2 of an A4 sheet: about 3.5\% less area.
14.3 mechanics in a parallel universe
In a universe parallel to ours, mechanics was first developed by Yitzhak Natan, who lived in the Land of Israel in 600 BCE. At that time, the base units were different from the SI units that we use today:
The base unit of length was the amah (cubit), equal to 52 \text{ cm}.
The base unit of mass was the shekel, equal to 14 \text{ g}.
The base unit of time was the chelek. A day has 24 hours, and an hour has 1080 chalakim (that’s the plural of chelek).
We’ll translate some quantities into the units that Yitzhak Natan was used to.
A chelek is three and a third seconds. We will use this in the next parts, together with 1 \text{ amah} = 0.52 \text{ m} and 1 \text{ shekel} = 0.014 \text{ kg}.
b. gravity in Natan’s units
The gravitational acceleration on Earth is g = 9.8 \text{ m/s}^2. What is its value in the units that Yitzhak Natan was used to?
Tipsetup: starting point, goal, conversion
An acceleration has the dimensions of length divided by time squared, L/T^2, so in Natan’s system it is measured in \text{amah/chelek}^2 (see dimensions). We convert the meters to amah and the seconds to chalakim. The time is squared, so its conversion factor is squared too.
In Natan’s units, g \approx 209 \text{ amah/chelek}^2.
c. atmospheric pressure
The atmospheric pressure is P_\text{atm} = 1.01 \cdot 10^5 \text{ Pa}. What is its value in the units that Yitzhak Natan was used to?
Tipsetup: starting point, goal, conversion
Pressure is force per area, and the pascal is 1 \text{ Pa} = 1 \text{ N/m}^2. A newton is 1 \text{ N} = 1 \text{ kg} \cdot \text{m/s}^2, so in base units
which has the dimensions M / (L \cdot T^2). So, in Natan’s system pressure is measured in \text{shekel}/(\text{amah} \cdot \text{chelek}^2). We convert each base unit separately: the kilograms to shekels, the meters to amah and the seconds to chalakim.
The meter is in the denominator of the pascal, so this time the factor of 0.52 \text{ m} goes on top, and the meter cancels. That’s the idea of the chain-link method: we choose the version of each factor that cancels the unit that we want to get rid of.
d. a light bulb
What is the power of a 30 \text{ W} light bulb in the units that Yitzhak Natan was used to? Recall that 1 \text{ W} = 1 \text{ J/s}.
Tipsetup: starting point, goal, conversion
Let’s write the watt in terms of base units. A joule is a newton times a meter, 1 \text{ J} = 1 \text{ N} \cdot \text{m} = 1 \text{ kg} \cdot \text{m}^2/\text{s}^2, so
In a world with different units, an ordinary light bulb would have a rather large number on its box: almost three hundred thousand!
14.4 rain in a box
a. water in the box
After a thunderstorm, a depth of 5 \text{ mm} of rain was measured in a box that was left outside. The box has a square base, with sides of 1 \text{ m}. How much water (in liters) collected in the box? Recall that a liter (L) is the volume of a cube with sides of 10 \text{ cm}.
Tipsetup: starting point, goal, conversion
The water in the box has the shape of a rectangular block: a base of 1 \text{ m} \times 1 \text{ m} and a height of 5 \text{ mm}. Its volume is the area of the base times the height. The liter is defined with centimeters, so let’s convert all the lengths to centimeters.
Starting point: the volume of the block of water, V = 1 \text{ m} \times 1 \text{ m} \times 5 \text{ mm}.
So one millimeter of rain is exactly one liter of water on each square meter of ground. This is why the amount of rain is usually reported in millimeters: it tells us the depth of the water layer, and also how many liters fell on every square meter.
b. rain on a lake
The same storm drops 5 \text{ mm} of rain over a lake with an area of 170 \text{ km}^2 (about the area of Lake Kinneret). Assume that all the water stays in the lake. How much water was added to the lake, in cubic meters?
Tipsetup: starting point, goal, conversion
We can use the result of part a as a conversion factor: every square meter receives 5 \text{ L}. Or, we can find the volume as the area times the depth, like before. Let’s use the second way, in meters.
Starting point: the area of the lake, 170 \text{ km}^2.
That’s 850{,}000 cubic meters, or 850 million liters, from a single storm! It’s about 340 Olympic swimming pools (each holds 2500 \text{ m}^3).
We can check this with part a: 1.7 \cdot 10^{8} \text{ m}^2 times 5 \text{ L/m}^2 is 8.5 \cdot 10^{8} \text{ L}, and 10^3 \text{ L} = 1 \text{ m}^3, so 8.5 \cdot 10^{5} \text{ m}^3. It’s the same result.