G water = mg = 0.6kg×10n/kg = 6n;

(2) the pressure of water on the bottom of the teapot:

P water = rho water GH =1.0×103kg/m3×10n/kg× 0.12m =1200pa;

(3) According to p= F S, the pressure of water on the bottom of the teapot is:

F water =p water S =1200 Pa× 40×10-4m2 = 4.8n;

(4) The pressure of the teapot on the desktop:

F desktop =G pot +G water =m pot g+G water = 400×10-3kg×10n/kg+6n =10n,

Pressure of teapot on desktop:

P Desktop = F Desktop S =10N40X10-4m2 = 2500pa;

(5) Communicator refers to a container with an open top and a communicated bottom. The spout and lid of the teapot are communicated with the atmosphere to ensure that the liquid level of the spout is always flat, so the knowledge of the correspondent is applied;

The small hole at the mouth of the kettle ensures that the air pressure in the kettle is equal to atmospheric pressure, and prevents the air pressure in the kettle from being too small and the water in the kettle from pouring out, so the knowledge of atmospheric pressure is also applied.

Answer: (1) The gravity of water is 6n;

(2) The pressure of water on the bottom of the teapot is1200 Pa; ;

(3) The pressure of water on the bottom of the teapot is 4.8N；;

(4) The pressure of the teapot on the desktop is10n; ;

(5) The pressure of the teapot on the desktop is 2500Pa；;

(6) Connector (or atmospheric pressure).