Physics, asked by Anonymous, 9 months ago

A block of mass 7kg and volume 0.07m^3 floats in a liquid of density 149kg/m^3. calculate:
(i) volume of block above the surface of liquid.
(ii)Density of block

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Answers

Answered by Anonymous

Answer:

weight of block = buoyancy force = weight of liquid displaced

Volume of liquid displaced = V m³

7 * 10 Newtons = V * 140 * 10 Newtons

V = 0.05 m³

Volume above the surface of liquid = 0.07 - 0.05 = 0.02 m³

==============================

mass of solid = 0.5 kg density = 5000 kg /m³

mass of liquid displaced by solid = volume of solid * density of liquid

= (0.5/5, 000) * 8, 000 = 0.8 kg

Buoyancy force on solid = 0.8 kg * 10 m/s² = 8 Newtons

When the solid is completely immersed in the liquid, the buoyancy force is more than its own weight. Hence, it will float immediately. It looks as if its apparent weight is negative and is -3 Newtons or -0.3 kg. ( 0.5g - 0.8g)

When it reaches the surface of the liquid, it will displace less volume of liquid such that buoyancy force = weight of solid. Then it floats. Hence apparent weight = 0.

hope it's help u

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Answered by Anonymous

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A block of mass 7kg and volume 0.07m^3 floats in a liquid of density 149kg/m^3. calculate:
(i) volume of block above the surface of liquid.
(ii)Density of block

please don't spam