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torisob [31]
3 years ago
10

A hollow cubical box is 0.221 m on an edge. This box is floating in a lake with 1/4 of its height beneath the surface. The walls

of the box have a negligible thickness. Water is poured into the box. What is the depth of the water in the box at the instant the box begins to sink
Physics
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

3/4 filled with water

Explanation:

x = Fraction of box filled

g = Acceleration due to gravity = 9.81 m/s²

\rho_w = Density of water

V = Volume of water

Weight of empty box is equal to the weight of the water displaced

W_b=\frac{1}{4}V\times \rho_wg

As all the forces are conserved

W_b+xV\rho_wg=V\rho_wg\\\Rightarrow \frac{1}{4}V\times \rho_wg+xV\rho_wg=V\rho_wg\\\Rightarrow \frac{1}{4}+x=1\\\Rightarrow x=1-\frac{1}{4}\\\Rightarrow x=\frac{2}{3}

So, the box starts to sink when the box is 3/4 filled with water

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Answer:

No. of Molecules = 18.06 x 10²³ molecules

No. of Atoms of P = 3 atoms

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Total No. of Atoms = 24 atoms

Explanation:

<u>FOR NUMBER OF MOLECULES</u>:

No.\ of\ Molecules = (No.\ of\ Moles)(Avigadro's\ Number)\\No.\ of\ Molecules = (3)(6.02\ x\ 10^{23}\ molecules})\\

<u>No. of Molecules = 18.06 x 10²³ molecules</u>

<u></u>

<u>FOR NUMBER OF P ATOMS</u>:

No.\ of\ Atoms\ of\ P = (No.\ of\ Moles)(N.\ of atoms of P)\\No.\ of\ Atoms\ of\ P = (3)(1\ atom})

<u>No. of Atoms of P = 3 atoms</u>

<u></u>

<u>FOR NUMBER OF Na ATOMS</u>:

No.\ of\ Atoms\ of\ Na = (No.\ of\ Moles)(N.\ of atoms of Na)\\No.\ of\ Atoms\ of\ Na = (3)(3\ atom})

<u>No. of Atoms of Na = 9 atoms</u>

<u></u>

<u>FOR TOTAL NUMBER OF ATOMS</u>:

Total\ No.\ of\ Atoms = (No.\ of\ Moles)(N.\ of\ atoms\ of\ Na + No.\ of\ atoms\ of\ P + No.\ of\ atoms\ of\ O)\\ Total\ No.\ of\ Atoms\ = (3)(3\ atoms + 1\ atom + 4\ atoms}) = (3)(8\ atoms)

<u>Total No. of Atoms = 24 atoms</u>

<u></u>

<u></u>

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