Why is mercury so dense? Go deeper

Why is mercury so dense? Go deeper

How to say, mercury is mercury Hg, its atomic mass is large, and much larger than the general liquid. Then the density is large!

The density is 0.6x10 ^ 3kg / m ^ 3, the side length is 10cm, the cube wood block is still on the water surface, calculate (1) the gravity of the wood block, (2) the buoyancy of the wood block
(3) The minimum vertical downward force to be applied on the upper surface of the block when the block is completely immersed in water?

(1) Block weight:
G wood = m wood, g = ρ wood, V wood, g = 0.6 × 103kg / m3 × (0.1M) 3 × 10N / kg = 6N,
(2) ∵ the wood floats,
The buoyancy of the block:
F = g = 6N,
(3) When the wood block is fully immersed, it is buoyant by water
F ′ = ρ water V drainage g = ρ water V wood g = 1 × 103kg / m3 × (0.1M) 3 × 10N / kg = 10N,
F ′ = f pressure + G wood,
F pressure = f floating '- G wood = 10n-6n = 4N
A: (1) the gravity of the block is 6N;
(2) The buoyancy of wood block is 10N;
(3) At least 4N force should be applied on the surface of the block to make the block immersed in water
analysis:
(1) Know the side length of the block, calculate the volume of the block, and calculate the weight of the block according to the formula of density and gravity;
(2) The weight of the block is calculated, and the buoyancy of the block is calculated by using the floating condition;
(3) Know the volume of the block (the volume of the whole immersion in the water), and calculate the buoyancy of the block according to Archimedes' principle. When the block is completely immersed in the water, the buoyancy of the block is equal to the sum of the gravity of the block and the vertical downward pressure, and then calculate the required pressure
Grasp these, is the key to learn physics!

The cube wood block with density of 0.6x10 ^ 3kg / m ^ 3 and side length of 10cm is still on the water surface

（1）V＝a3＝（0.1m）3＝0.001m3.
m＝ρV＝0.6x10^3kg/m^3x0.001m3＝0.6kg
G wood = mg = 0.6kgx10n / kg = 6N
(2) Because the cube block is still on the water surface, f floating = g, so f floating = 6N

There is a rectangular wood block with length of 20cm, width of 10cm and height of 5cm. The density is 0.4 times the power of ten. If it floats on the water, what is the buoyancy of the wood block?

If the object floats, it means that the gravity and buoyancy of the object are balanced, equal in size and opposite in direction
V = 20 * 10 * 5 = 1000cm ^ 3
M = 1000 * 0.4 = 400g = 0.4KG
G = 9.8 * 0.4 = 3.92n = f buoyancy

The buoyancy of 7.9g solid iron ball and 8.9g solid copper ball in water and mercury was calculated;
The density of mercury is 13.6g/cm and copper is 8.9g/cm;

The density of the solid iron ball floating in mercury is less than that of mercury. The buoyancy of the solid iron ball floating in mercury is f Fe = G1 = m1g = 0.0079kg * 10N / kg = 0.079n. The buoyancy of the solid copper ball floating in mercury is f Cu = G2 = m2g = 0.0089kg * 10N / kg = 0.089n. The density of the solid iron ball floating in mercury is greater than that of water

There are three pieces of equal mass of copper, iron and aluminum, which are put into a pot of mercury. Their buoyancy is F1, F2 and F3. Try to compare the three forces

Because the density of copper, iron and aluminum is less than that of mercury, the three equal masses of copper, iron and aluminum float on the surface of mercury, so their buoyancy is equal to gravity, that is, F1 = F2 = F3

The buoyancy of copper, iron and aluminum solid metal blocks with the same mass after being put into mercury is F1, F2 and F3, respectively
A. F1 ＞ F2 ＞ F3B. F1 = F2 = F3C. F1 ＜ F2 ＜ F3D

∵ three metal blocks have the same mass ∵ three metal blocks have the same gravity g ∵ copper, iron and aluminum blocks are solid, and their density is less than that of mercury ∵ when they are put into mercury, they will float, ∵ F1 = F2 = F3 = g, so B is selected

What is the buoyancy of a 10 cm * 3 iron ball when it is put into mercury at rest?
Such as the title

First of all, you should understand that the density of mercury is much higher than that of iron, so the iron ball floats on the surface of mercury, so the buoyancy is equal to the gravity of the iron ball

When a small solid iron ball with a volume of 10 cm ^ 3 is put into water, the buoyancy of the ball at rest is 0_ N. When you put it in mercury, the buoyancy at rest is u n

F1 = 1000kg / m3 × 10N / kg × 0.0000001 = 0.001N because the density of mercury is higher than that of iron, so the iron ball floats on the mercury. F2 = g = 0.0000001 × 7800kg / m3 = 0.00075n

The buoyancy of a copper ball immersed in water is F1, and that of a copper ball immersed in kerosene is F2
A. F1 ＞ f2b. F1 ＜ F2C. F1 = F2D

The density of water is greater than that of kerosene, and the volume of copper ball remains unchanged. According to the formula F = ρ GV, the buoyancy of copper ball in water is greater than that in kerosene