When a solid metal ball with a weight of 38n and a volume of 5 × 10 - is immersed in a sleeping container, what is the weight of the overflow water and what is the buoyancy of the metal ball

When a solid metal ball with a weight of 38n and a volume of 5 × 10 - is immersed in a sleeping container, what is the weight of the overflow water and what is the buoyancy of the metal ball

I'm talking nonsense upstairs
F = row G
Because it is immersion, the volume of the metal ball = the volume of the drained water. The gravity of water can be calculated from the volume of the drained water. The gravity of water is equal to the buoyancy of the metal ball
F floating = 5 × 10 - (I don't know what unit you wrote) * 1kg / m ^ 3 (density of water) * 9.8n/kg

A solid metal ball with a weight of 38n and a volume of 5 × 10-4 Power M & # 179; is hung under a spring dynamometer, and then the metal ball is immersed in a container full of water. What is the buoyancy of the metal ball? What is the indication of the spring dynamometer? (G is 10N / kg)

A solid metal ball with a weight of 38 N and a volume of 5 × 10 ＾ - 4 m3 was immersed in a container full of water, and the weight of the beneficial water was
The buoyancy of the metal ball is————

The amount of water spilled has nothing to do with the weight of the metal ball
It's only about volume
The density of water is 1000 kg / m3
The volume of spilled water is 5 × 10-4 cubic meters
So weight = 1000 × 5 × 10 ^ - 4
=0.5kg
Buoyancy = weight of discharged water = 0.5 × 10 = 5N

A metal ball with a mass of 5 kg or 4 kg is immersed in a water tank filled with water, and the volume of spilled water is 2x10-3 square meters (G is 10N / kg)
(1) The density of a metal ball
(2) The buoyancy of a metal ball
(3) The amount of pressure the metal ball exerts on the lower part of the tank

ρ=m/v=5.4kg/2×10^-3 m3 =2700kg/m3
F = ρ GV = 10 ^ 3kg / m3 × 10N / kg × 2 × 10 ^ - 3m3 = 20n
F pressure = G-F floating = 5.4kg × 10N / kg - 20n = 34n