p01_044

p01_044 - 2 gal − 1 = 2 17 × 10 − 3 gal ft 2 which is...

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44. (a) Using Appendix D, we have 1 ft = 0 . 3048 m, 1 gal = 231 in . 3 ,and1in . 3 =1 . 639 × 10 2 L. From the latter two items, we ±nd that 1 gal = 3 . 79 L. Thus, the quantity 460ft 2 / gal becomes µ 460 ft 2 gal ¶µ 1m 3 . 28 ft 2 µ 1 gal 3 . 79 L =11 . 3m 2 / L . (b) Also, since 1 m 3 is equivalent to 1000 L, our result from part (a) becomes µ 11 . 3m 2 L ¶µ 1000 L 1m 3 =1 . 13 × 10 4 m 1 . (c) The inverse of the original quantity is (460 ft
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Unformatted text preview: 2 / gal) − 1 = 2 . 17 × 10 − 3 gal / ft 2 , which is the volume of the paint (in gallons) needed to cover a square foot of area. From this, we could also ±gure the paint thickness (it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part (b))....
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This note was uploaded on 04/08/2008 for the course PHY 1017 taught by Professor Hunt during the Spring '08 term at RIT.

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