74.8k views
1 vote
Oil having a density of 921 kg/m3 floats on water. A rectangular block of wood 4.18 cm high and with a density of 969 kg/m3 floats partly in the oil and partly in the water. The

oil completely covers the block. How far below the interface between the two liquids is the bottom of the block?
Answer in units of m

User Aniyah
by
6.9k points

2 Answers

5 votes

Answer:

The buoyant force on the block is equal to the weight of the displaced fluid, which in this case is the combined weight of the oil and water displaced by the block. Let's call the height of the block that is submerged in the water "h" and the height of the block that is submerged in the oil "d".

The buoyant force on the block is given by:

F_b = (density of water)(gravity)(volume of water displaced by the block) + (density of oil)(gravity)(volume of oil displaced by the block)

The weight of the block is given by:

F_g = (density of wood)(gravity)(volume of the block)

Since the block is floating, the buoyant force is equal to the weight of the block, so we can set F_b = F_g:

(density of water)(gravity)(h)(cross-sectional area of the block) + (density of oil)(gravity)(d)(cross-sectional area of the block) = (density of wood)(gravity)(height of the block)(cross-sectional area of the block)

Simplifying and solving for d, we get:

d = [density of wood / density of oil - (density of water / density of oil) * h] * height of the block

Plugging in the given values, we get:

d = [969 kg/m3 / 921 kg/m3 - (1000 kg/m3 / 921 kg/m3) * 0.0418 m] * 0.0418 m

d ≈ 0.0246 m

Therefore, the bottom of the block is about 0.0246 meters (or 2.46 centimeters) below the interface between the oil and water.

Step-by-step explanation:

User JamesTBennett
by
7.2k points
5 votes

Answer:

bc....

Step-by-step explanation:

this is for ,,,,,

User Kalahari
by
7.1k points