Question

in liquid]
9. A block of wood is floating on water with its
dimensions 50 cm x 50 cm x 50 cm inside water.
Calculate the buoyant force acting on the block. Take
g = 9.8 N kg-1.​

Answer 1

Answer:

Density of water, \rho = 1000 \ kg \ {m}^{-3}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\times 50 \times 50 \times {10}^{-6}\ {m}^{3} = 0.125\ {m}^{3}50×50×50×10

−6

m

3

=0.125 m

3

Weight of water displaced = 0.125 \times 1000 \times 9.8 = 1225\ N0.125×1000×9.8=1225 N

X = 1225X=1225

Answer 2

Answer:

Explanation:

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