CH-01 - Chapter 1 1 Using the given conversion factors we...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 1 1. Using the given conversion factors, we find (a) the distance d in rods to be ( 29 ( 29 4.0 furlongs 201.168 m furlong 4.0 furlongs = 160 rods, 5.0292 m rod d = = (b) and that distance in chains to be ( 29 ( 29 4.0 furlongs 201.168 m furlong 40 chains. 20.117 m chain d = = 2. The conversion factors 1 gry 1/10 line = , 1 line=1/12 inch and 1 point = 1/72 inch imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry 2 = (0.60 point) 2 = 0.36 point 2 , which means that 2 2 0.50 gry = 0.18 point . 3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2). (a) Since 1 km = 1 × 10 3 m and 1 m = 1 × 10 6 μ m, ( 29 ( 29 3 3 6 9 1km 10 m 10 m 10 m m 10 m. = = = μ μ The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 10 9 μ m. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10 - 2 m, ( 29 ( 29 2 2 6 4 1cm = 10 m = 10 m 10 m m 10 m. - - = μ μ We conclude that the fraction of one centimeter equal to 1.0 μ m is 1.0 × 10 - 4 . (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, ( 29 ( 29 6 5 1.0 yd = 0.91m 10 m m 9.1 10 m. = × μ μ 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 1 4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain ( 29 1 inch 6 picas 0.80 cm = 0.80 cm 1.9 picas. 2.54 cm 1 inch    (b) With 12 points = 1 pica, we have ( 29 1 inch 6 picas 12 points 0.80 cm = 0.80 cm 23 points. 2.54 cm 1 inch 1 pica       5. Various geometric formulas are given in Appendix E. (a) Substituting R = × = × - 6 37 10 10 6 37 10 6 3 3 . . m km m km c hc h into circumference = 2 π R , we obtain 4.00 × 10 4 km. (b) The surface area of Earth is ( 29 2 2 3 8 2 4 4 6.37 10 km 5.10 10 km . A R = π = π × = × (c) The volume of Earth is ( 29 3 3 3 12 3 4 4 6.37 10 km 1.08 10 km . 3 3 V R π π = = × = × 6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1 fanega = cahiz, or 8.33 × 10 - 2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = cahiz, or 2.08 × 10 - 2 cahiz. Continuing in this way, the remaining entries in the first column are 6.94 × 10 - 3 and 3 3.47 10 - × . (b) In the second (“fanega”) column, we similarly find 0.250, 8.33 × 10 - 2 , and 4.17 × 10 - 2 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get = 0.500 for the last entry. 2
Image of page 2
(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94 × 10 - 3 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 × 10 - 2 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m 3 or 55501 cm 3 . Thus, 7.00 almudes = fanega = (55501 cm 3 ) = 3.24 × 10 4 cm 3 .
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern