Health pot p 1
Solution: the pressure of water at the bottom of the teapot p 1=pgh= 1000kg/m? x 10N/kgx 0. 12m = 1200 pa

The pressure of water on the bottom of teapot F 1=p 1S pot = 1200Pax4× 10 negative cubic meter = 4.8n.

The pressure of the teapot on the desktop F2=G pot +G water =m pot g+m water g = 0.4kgx10n/kg+0.6kgx10n/kg =10n.

The pressure of the teapot on the desktop p2=F2/S pot = 10N/4× 10 negative cubic meter = ㎡=2.5x 10/0? protactinium