Question

A seal with a mass of 130 kg floats on a slab of ice in the Arctic Ocean. What minimum volume must the slab have in order that the seal not get its skin wet with the seawater? In other words, the top surface of the ice slab is exactly at the same level as the seawater. Use 1025 kg/m3 for the density of seawater and 920 kg/m3 for the density of the ice slab. a) 1.24 m3 b) 1.01 m3 c) 1.43 m3 d) 1.33 m3 e) 1.14 m3

Answer 1

2 answers · Physics

The density difference between water and ice = 1025 kg/m^3 - 920 kg/m^3 = 105 kg/m^3

130 kg / 105 kg/m^3 = 1.24 m^3

Ricki · 3 years ago

Comment

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The buoyanr force must equal the weight of the slab and the seal. The buoyant force, FB ,
equals the weight of sea water displaced by the seal and the slab.
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FB = buoyant force = ( V slab ) ( dsw ) ( g )

FB = Wseal + W slab = ( mseal ) ( g ) + ( Vslab ) ( dice ) ( g )

( Vslab ) ( dsw ) ( g ) = ( mseal ) ( g ) + ( Vslab ) ( dice ) ( g )

( Vslab ) ( dsw - dice ) = mseal

Vslab = ( mseal ) / ( dsw - dice )

Vslab = ( 130 kg ) / ( 1025 kg/m^3 - 920 kg/m^3 )

Vslab = ( 130 kg ) / ( 105 kg/m^3 ) = 1.238 m^3 = 1.24 m^3<---------------------

Hence option A is correct.

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