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2 years ago

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An aluminum block has a mass of 0.25 kg and a density of 2700 kg/m3. a) Calculate the volume of the aluminium block. b) Determin

e the buoyant force exerted on this block when it is completely submerged in water of density 1000 kg/m3

1 answer:

Hatshy [7]2 years ago

7 0

Answer:

a

The volume of the aluminum is $V = 9.25*10^{-5}m^3}$

b

The buoyant force exerted is $F_B =0.9N$

From the question we are told that

The mass of the aluminum is $M_A = 0.25Kg$

The density of the aluminum is $\rho_A = 2700 kg/m^3$

Generally volume is mathematically represented as

$V = \frac{M_A}{\rho_A}$

Substituting values accordingly

$V =\frac{0.25}{2700} =9.25*10^{-5}m^3$

Generally buoyant force is mathematically represented as

$F_B = V \rho_wg$

Where V is the volume of the aluminum $=9.25*10^{-5}m^3$

$\rho_w$ is the density of water $=1000kg/m^3$

g is the acceleration due to gravity = $=9.8m/s^2$

Substituting these values accordingly we have

$F_B = 9.25*10^{-5} * 1000 * 9.8$

$=0.9N$

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