Question #7937

A hollow brass tube (diameter = 3.40 cm) is sealed at one end and loaded with lead shot to give a total mass of 0.191 kg. When the tube is floated in pure water, what is the depth, z, of its bottom end?

Expert's answer

Total Weight = Buoyant Force

mass * g = Buoyant Force

mass = 0.191 kg.

0.191 kg* 9.807= Specific Weight * Volume Submerged

---> Volume Submerged = Area * Depth

---> Specific Weight = p * g

--->p=1 g/cm3 = 1000kg / m3

--->g=9.807 m/s2

--->Area = pi *(diameter / 2)^2

0.191 kg= p * g* Area* Depth

0.191 kg* 9.807m/s2 = (1000kg / m3)(9.807 m/s2) * (pi* (1.7/100)^2 m) * Depth

Depth = Final Answer = 21,048cm

mass * g = Buoyant Force

mass = 0.191 kg.

0.191 kg* 9.807= Specific Weight * Volume Submerged

---> Volume Submerged = Area * Depth

---> Specific Weight = p * g

--->p=1 g/cm3 = 1000kg / m3

--->g=9.807 m/s2

--->Area = pi *(diameter / 2)^2

0.191 kg= p * g* Area* Depth

0.191 kg* 9.807m/s2 = (1000kg / m3)(9.807 m/s2) * (pi* (1.7/100)^2 m) * Depth

Depth = Final Answer = 21,048cm

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