Asked by Peter
determine the submerged depth of a cube of steel 0.3m on each side floating in mercury given the specific gravities of steel and mercury as 7.8 and 13.6 respectively.
Answers
Answered by
MathMate
Principle of floatation:
A floating object displaces its own weight of the fluid in which it floats.
Volume of steel
=V<sub>s</sub>
Mass of steel
M<sub>s</sub>
= ρ<sub>s</sub>*V<sub>s</sub>
Volume of mercury displaced, V<sub>m</sub> = M<sub>s</sub> / ρ<sub>m</sub>
Percentage of volume displaced
=V<sub>m</sub> / V<sub>s</sub>
If the steel block floats with bottom face parallel to surface of mercury, submerged depth equals height of block multiplied by percentage of volume displaced.
It is slightly less than two-thirds of the height of the block.
A floating object displaces its own weight of the fluid in which it floats.
Volume of steel
=V<sub>s</sub>
Mass of steel
M<sub>s</sub>
= ρ<sub>s</sub>*V<sub>s</sub>
Volume of mercury displaced, V<sub>m</sub> = M<sub>s</sub> / ρ<sub>m</sub>
Percentage of volume displaced
=V<sub>m</sub> / V<sub>s</sub>
If the steel block floats with bottom face parallel to surface of mercury, submerged depth equals height of block multiplied by percentage of volume displaced.
It is slightly less than two-thirds of the height of the block.
Answered by
Anonymous
1.7199m
Answered by
Anonymous
0.0155
Answered by
Mandefro
.172
Answered by
Bot
Sorry, I'm not sure what those numbers refer to. Can you please provide more context or information so I can assist you better?
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