(1)∵ The block floats,

∴F float = gram wood,

∫F float = ρ water v row g, g wood = ρ wood v wood g,

∴ ρ water v row g=ρ wood v wood g

∫ 1/5 is exposed to water,

∴V line =4/5 V wood,

∴ ρ wood =4/5 ρ water = 4/5×/kloc-0 /×103kg/m3 = 0.8×103kg/m3;

(2) As shown in the figure, when the string is broken, F floats' +F max =G wood,

At this point, let the volume of boiling water discharged from the wood block be V rows, and then:

ρ water v row' g+F max = ρ wood v wood g,

That is, 1× 10 3kg/m 3× V row '×10n/kg+5n = 0.8×103kg/m3× 0.1m3× 65438.

Solution:

Row v' = 3×10-4m3;

(3)△V row =V row -V row' =4/5 V wood-3×10-4m3 = 4/5× 0.1m3-3×10-4m3 = 5× 65438+.

△h=△V platoon/s = 5×10-4m3/0.02m2 = 0.025m,

△p=ρ water g△ h =/kloc-0 /×103kg/m3×10n/kg× 0.025m = 250pa,

In other words, the water pressure at the bottom of the container increased by 250 Pa. ..

Answer: (1) The density of wood blocks is 0.8×10.3 kg/m3;

(2) When the string is broken, the valve closes instantly, and the open water discharged by the wood block is 3×10-4m3;

(3) When the wooden block floats again after the rope is broken, the water pressure at the bottom of the container increases by 250Pa. Compared with the moment before the string is broken.