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Unformatted text preview: 45. (a) If the volume of the car below water is V1 then Fb = ρwV1g = Wcar, which leads to
V1 = ( ) (1800 kg ) 9.8 m s2
Wcar
=
= 1.80 m3 .
3
2
ρ w g 1000 kg m 9.8 m s ( )( ) (b) We denote the total volume of the car as V and that of the water in it as V2. Then
Fb = ρ wVg = Wcar + ρ wV2 g
which gives
V2 = V − Wcar
1800 kg
= 0.750 m3 + 5.00 m3 + 0.800 m3 −
= 4.75 m3 .
3
1000 kg m
ρw g ( ) ...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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