When a solid aluminum ball with a mass of 2.7kg is put into mercury and water respectively, what is the buoyancy of the aluminum ball? (g = 10N / kg)

When a solid aluminum ball with a mass of 2.7kg is put into mercury and water respectively, what is the buoyancy of the aluminum ball? (g = 10N / kg)

From ρ = m / V
The solid aluminum ball with M = 2.7kg is obtained, and the volume is v ball = m / ρ = 1x10 ^ - 3 m ^ 3
Because ρ aluminum is less than ρ mercury, the aluminum ball floats in mercury
Then f = g = mg = 27N
Because ρ aluminum is greater than ρ water, the aluminum ball sinks in water
F floating '= ρ water GV ball = 10N
For reference only!

The buoyancy of a solid metal ball in water 'after resting in mercury' is 0.8N and 1n respectively. The mass' volume 'density of the metal ball can be calculated

1, floating in mercury
F = g
G = 1n
G matter = m matter G
M = 0.1kg
2. Immerse in water
F floating = ρ liquid GV discharge
V row = V object
V = 0.8/10000
=0.8*10^-4m^3
=80cm^3
3,
G matter = ρ matter GV matter
ρ = 1.25 * 10 ^ 3kg / m ^ 3

When a solid object with a volume of 2 and a density of 0.5 × 10 kg / M is put into water, how much buoyancy does it receive when it is still?
What is the state of the object at this time? (fill in "floating", "floating", or "sinking at the bottom"). If it is put into a liquid with a density of 0.4 × 10 ~ (- 179); kg / M ~ (- 179), what is the buoyancy n when it is still?

The density of solid objects with the density of 0.5 × 10 ~ (- 179) kg / M ~ (- 179) is less than that of water, so they float in water
F floating = g matter = m matter g = ρ matter V matter g solid = 500kg / m3 * 2m3 * 10N / kg = 10000n
Sink into a liquid with a density of 0.4 × 10 ~ (- 179) kg / M ~ (- 179)
F floating = g row = m row g = ρ V row g = 400kg / m3 * 2m3 * 10N / kg = 8000n