A Rectangular wooden block of thickness 5cm and base area 20cm2 float in water with 3.

Question

A Rectangular wooden block of thickness 5cm and base area 20cm2 float in water with 3.5cm of thickness below the water surface.if the side are equally above water calculate the density of the wooden block

5cm of thickness below the water surface.

Answer 1

The volume of the wooden block can be calculated by multiplying the base area by the thickness:
Volume = Base Area x Thickness
Volume = 20 cm^2 x 5 cm
Volume = 100 cm^3

Since the entire block is submerged in water, the buoyant force is equal to the weight of the water displaced by the block:
Buoyant Force = Weight of Water Displaced
Buoyant Force = Density of Water x Volume of Water Displaced x Acceleration due to Gravity
Buoyant Force = 1000 kg/m^3 x 20 cm^2 x 3.5 cm x 0.01 m/cm x 9.81 m/s^2
Buoyant Force = 6.93 N

The weight of the block is equal to its density times its volume times the acceleration due to gravity:
Weight = Density x Volume x Acceleration due to Gravity
Weight = Density x 100 cm^3 x 0.01 m/cm x 9.81 m/s^2

Since the wooden block is in equilibrium, the weight of the block must be equal to the buoyant force:
Density x 100 cm^3 x 0.01 m/cm x 9.81 m/s^2 = 6.93 N

Solving for density, we get:
Density = 6.93 N / (100 cm^3 x 0.01 m/cm x 9.81 m/s^2)
Density ≈ 7.1 x 10^3 kg/m^3

Therefore, the density of the wooden block is approximately 7.1 x 10^3 kg/m^3.