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Changning district 20 15 bimodule mathematics
① It can be obtained from ρ=mV.

M armour = ρ metal V armour = 5.0×103kg/m3× 2×10-3m3 =10kg;

②p water = rho water GH =1.0×103kg/m3× 9.8n/kg× 0.1m = 980pa;

(3) When d ≤ h, the surface pressure of liquid on the nail bottom is p nail bottom = ρ gd;

The pressure at the bottom of liquid against B is P =ρg(H+dS3S).

P nail bottom p capacity bottom = ρ gdρ g (h+ds3s) = 3d3h+d;

When d > h, the pressure of liquid on the lower surface of nail is p nail bottom = ρ GD;

The bottom pressure of liquid to b is p =ρg(H+hS3S),

P nail bottom p capacity bottom = ρ gdρ g (h+hs3s) = 3d3h+h 。

Answer: ① The mass of A is 10kg. ..

② The water pressure at the depth of 0.1m is 980Pa;;

(3) when d ≤ h, the bottom of p nail is equal to 3d ≤ h+d;

When d > h, P nail bottom and P volume = 3d3h+h. 。