Question

A small block of wood of mass 9 kg is floating on water. Find the force of buoyancy acting on the block. (g = 10 ms2)​

Answer 1

Answer:

1225 N

Explanation:

ANSWER

Density of water, ρ=1000 kg m

−3

According to Archimedes principle, the buoyant force acting on an object is equal to the weight of the liquid displaced.

Volume of water displaced = 50×50×50×10

−6

m

3

=0.125 m

3

Weight of water displaced = 0.125×1000×9.8=1225 N

X=1225

Answer 2

Explanation:

Given A small block of wood of mass 9 kg is floating on water. Find the force of buoyancy acting on the block.  

• Density of water = 1000 kg / m^3
• Mass = 9 kg
• Now we have density = mass / volume.
•           So volume = mass / density
•                           = 9 kg / 1000 kg / m^3
•                              = 9 / 1000 m^3
• Now force of buoyancy is given by F = ρ x g x V  
•                                                         F = 1000 x 10 x 9 / 1000
•                                                         F = 90 N
• So force of buoyancy acting on the block is 90 N

Reference link will be

https://brainly.in/question/1095833