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Hatshy [7]
3 years ago
13

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
Physics
1 answer:
Hatshy [7]3 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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