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Unformatted text preview: 34. (a) We find the volume in cubic centimeters (193 gal) 231 in3 1 gal 2.54 cm 1 in
3 = 7.31 105 cm3 and subtract this from 1 106 cm3 to obtain 2.69 105 cm3 . The conversion gal in3 is given in Appendix D (immediately below the table of Volume conversions). (b) The volume found in part (a) is converted (by dividing by (100 cm/m)3 ) to 0.731 m3 , which corresponds to a mass of 1000 kg/m3 0.731 m2 = 731 kg using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731 kg = 4.06 105 min 0.0018 kg/min which we convert to 0.77 y, by dividing by the number of minutes in a year (365 days)(24 h/day)(60 min/h). ...
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This note was uploaded on 06/11/2008 for the course PHY 034 taught by Professor Furlan during the Spring '08 term at UniFor.
 Spring '08
 FURLAN

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