What's the volume of water required inside the lock if

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The discussion revolves around calculating the volume of water required inside a lock with dimensions of 40m long, 3.5m high, and 6m wide, while accounting for a 0.5m height allowance. The total weight of the ship and its load is 700 tons, which displaces 700 cubic meters of water. The total volume of the lock, minus the displaced water, is calculated to be 700 cubic meters. The conversation also explores the implications of the ship emptying its load, questioning how this would affect the displaced volume and the water level in the lock. The calculations and principles applied are based on Archimedes' principle and the mass of water displaced.
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What's the volume of water required inside the lock if...

Homework Statement



The dimension of the lock is:40 long x 3.5 height x 6 width
allowing 0.5 height.

The ship weights 400T plus 300T(LOAD)
total= 700T

whats the amount of water used per lock allowing allowing 0.5 m from the height?

What if the ship empties its load on the lock ? the density of the load is 755kg/m^3.

Homework Equations



Archimedes principle
Mass of water displaced

The Attempt at a Solution



Mass of water displaced by the boat mass = 700T
VOLUME = 700M^3
Total volume inside the lock allowing 0.5

(3x40x6)m^3 - 700M^3

Is that correct ?

How about if the ship empties its load on the lock ?
 
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davekardle said:

Homework Statement



The Attempt at a Solution



Mass of water displaced by the boat mass = 700T
VOLUME = 700M^3
Total volume inside the lock allowing 0.5

(3x40x6)m^3 - 700M^3

Is that correct ?
Yes.

How about if the ship empties its load on the lock ?

The new displaced volume of water will be how much? So the level of the lock rises by how much?
 


Duplicate posting in several other forums.
 
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