Pressure of water on the bottom of the cup: f = PS =1.0×103 pa× 5.0×10-3m2 = 5n;
(2) The volume of boiled water discharged by the fitness ball: row V =Sh liter = 5.0×10-3m2× 2.2×10-2m =1./kloc-0 /×10-4m3,
Buoyancy of fitness ball: F float = ρ water gV row =1.0×103kg/m3×10n/kg×/kloc-0 /×10-4m3 =/kloc.
Answer: (1) The pressure of water on the bottom of the cup is 5N, and the pressure is 1.0× 103 Pa.
(2) The buoyancy of the fitness ball is1.1n.
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