How much buoyancy does a 2.7kg aluminum block bear when it is completely immersed in water? (ρ al = 2.7 × 103 kg / m3, ρ water = 1.0 × 103 kg / m3)
(1) From ρ = MV, the volume of aluminum block can be obtained as follows: v = m, ρ = 2.7kg2.7 × 103kg / m3 = 1 × 10-3m3, (2) ∵ when the aluminum block is immersed in water, ∵ V row = v = 1 × 10-3m3, f float = ρ water, GV row = 1 × 103kg / m3 × 9.8n/kg × 1 × 10-3m3 = 9.8N
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