559_Physics ProblemsTechnical Physics

559_Physics ProblemsTechnical Physics - Chapter 19 P19.46...

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Chapter 19 561 P19.46 At 0°C, 10.0 gallons of gasoline has mass, from ρ = m V mV == F H G I K J = 730 10 0 000380 100 27 7 kg m gal m gal kg 3 3 ej bg . . . . The gasoline will expand in volume by ∆∆ VV T i × ° ° ° = −− β 9 60 10 10 0 20 0 0 0 0 192 41 .. . . . C gal C C gal a f At 20.0°C, 10 192 27 7 gal kg = 10 0 27 7 27 2 . gal kg 10.0 gal 10.192 gal kg = F H G I K J = The extra mass contained in 10.0 gallons at 0.0°C is 27 7 27 2 0 523 ... kg kg kg −= . P19.47 Neglecting the expansion of the glass, h V A T h = = × ×° ° = π 4 3 3 3 2 0250 200 10 182 10 300 355 . . . cm 2 cm CC c m af FIG. P19.47 P19.48 (a) The volume of the liquid increases as VVT i A = . The volume of the flask increases as
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