A cube solid iron block with a mass of 7.9kg is placed in the center of a horizontal tabletop with an area of 0.5m2. What is the pressure of the iron block on the tabletop? (the density of the iron is 7.9 × 10 & # 179; kg / M & # 179;, g = 10N / kg)

A cube solid iron block with a mass of 7.9kg is placed in the center of a horizontal tabletop with an area of 0.5m2. What is the pressure of the iron block on the tabletop? (the density of the iron is 7.9 × 10 & # 179; kg / M & # 179;, g = 10N / kg)

V=M/P=7.9kg/7.9x10^3kg/m3=0.001m3
The side length of cube is 0.1M
Square bottom area 0.1x0.1 = 0.01m2
p=F/S=7.9kgx10N/kg/0.01m2=7900pa

The homogeneous cube wood block with 10 cm side length is still on the water surface, and the depth of its lower surface from the water surface is 8 cm
1. What is the pressure of water on the lower surface of cube?
2. How buoyant is the block?
3. What's the density of the block? (there should be a process. Good answer, 20 points)

1 according to P = pGH = 1000 * 9.8 * 0.08 = 784 PA
The water pressure on the lower surface of a cube is equal to buoyancy
F = PS = 784 * 0.1 * 0.1 = 7.84 n
P = m / v = 7.84 / 9.8 / (0.1 * 0.1 * 0.1) = 0.8 * 10 ^ 3 (kg / m3)

There is a cube with density of 0.6 × 10 ᦉ kg / m ᦉ 179; and side length of 10 cm
Slowly press the wood block into the water by hand. When all the wood blocks are just submerged in the water, how much pressure does the hand put on the wood block?

According to the stress analysis: f pressure + g weight = f floating
F floating = (density of water) * (acceleration of gravity) * (volume of drainage, i.e. volume of object)
G weight = (density of wood) * (volume of wood) * (acceleration of gravity)
Finishing: f pressure = (density of water - density of wood) * (volume of wood) * (acceleration of gravity)
Remember: units must be unified