Arectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. suppose that a 410-kg crate of auto parts is loaded onto the barge.
part a) how much deeper does the barge float?

The depth in which the barge floats when the crate was loaded on it is 0.033 m.

The given parameters;

The volume of the barge displaced by when the crate was loaded is calculated as follows;

The depth in which the barge floats when the crate was loaded on it is calculated as follows;

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Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float=

Density of water=

Weight of water displaced by barge=Buoyant force=-Weight of horse

1 m=100 cm

cm

Hence, the depth of barge float=3 cm

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: