# ch01-p032 - × 10 − 6 chaldron(c In the third...

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32. Table 7 can be completed as follows: (a) It should be clear that the first column (under “wey”) is the reciprocal of the first row – so that 9 10 = 0.900, 3 40 = 7.50 × 10 2 , and so forth. Thus, 1 pottle = 1.56 × 10 3 wey and 1 gill = 8.32 × 10 6 wey are the last two entries in the first column. (b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 caldron (that is, the entries along the “diagonal” in the table must be 1’s). To find out how many chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 1 12 chaldron = 1 bag. Thus, the next entry in that second column is 1 12 = 8.33 × 10 2 . Similarly, 1 pottle = 1.74 × 10 3 chaldron and 1 gill = 9.24
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Unformatted text preview: × 10 − 6 chaldron. (c) In the third column (under “bag”), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1 pottle = 2.08 × 10 − 2 bag, and 1 gill = 1.11 × 10 − 4 bag. (d) In the fourth column (under “pottle”), we find 1 chaldron = 576 pottle, 1 bag = 48 pottle, 1 pottle = 1 pottle, and 1 gill = 5.32 × 10 − 3 pottle. (e) In the last column (under “gill”), we obtain 1 chaldron = 1.08 × 10 5 gill, 1 bag = 9.02 × 10 3 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill. (f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since each bag is 0.1091 m 3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m 3 ....
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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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