A rectangular 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?

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lublana lublana · Answer 1 · 2020-01-17T10:07:59+01:00

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=[tex]l\times b\times h=5.2\times 2.4\times h=12.48h[/tex]

Density of water=[tex]10^3kg/m^3[/tex]

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

[tex]Volume\;of\;water\times density\;of\;water\times g=410\times g[/tex]

[tex]12.48h\times 1000=410[/tex]

[tex]h=\frac{410}{12.48\times 1000}=0.03 m[/tex]

1 m=100 cm

[tex]0.03 m=0.03\times 100=3[/tex]cm

Hence, the depth of barge float=3 cm

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-18T18:29:06+01:00

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;

[tex]W_n = \rho Vg\\\\V = \frac{W_n}{\rho g} \\\\V = \frac{410 \times 9.8}{1000 \times 9.8} \\\\V = 0.41 \ m^3[/tex]

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

[tex]lwd = 0.41\\\\d = \frac{0.41}{lw} \\\\d = \frac{0.41}{5.2 \times 2.4} = 0.033 \ m[/tex]

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