Ahollow cubical box is 0.766 m on an edge. this box is floating in a lake with 1/6 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?

0.637 m

Explanation:

Data provided:

Side of the hollow cubical box = 0.766 m

Total volume of the box, V = (0.766)³ = 0.449 m³

Depth under the water, d = 1/6 of the total depth = 0.766/6

Now, according to the Archimedes principle,

Weight of box = weight of water displaced( = density × volume of water displaced)

now, the volume of water displaced = volume of box submerged

= 0.766² × (0.766/6) = 0.0749 m³

also, density of water = 1000 kg/m³

thus,

weight of the box = 0.0749 m³ × 1000 = 74.9  kg

now,

When water is poured such that the total weight of water equals the weight of the remaining volume of box, the box will begin to sink.

therefore,

Remaining volume of the box = Total volume - submerged volume

or

Remaining volume of the box = 0.449 m³ - 0.0749 m³ = 0.374 m³

Thus, the height of 0.374 m³ of water when box starts to sink

volume = area of the box × depth of the water

or

0.374 m³ = 0.766² × depth of the water

or

depth of the water = 0.637 m

answer would be d.) i hope this

explanation:

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length