Consider a large cubic block of ice floating in seawater. the specific gravities of ice and seawater are 0.92 and 1.025, respectively. if a 25cm-high portion of the ice extends above the surface of the water, determine the height of the ice block below the surface?

219.04 cm

Explanation:

Given:

The specific gravity of ice, = 0.92  

The specific gravity of seawater,   = 1.025

The height of the block above the water surface, h = 25 cm

Now,

Density of sea water = Density of water × = 1000 kg/m³ × 1.025 = 1025 kg/m³

Density of ice = Density of water × = 1000 kg/m³ × 0.92 = 920 kg/m³

Now,

Let 'A' be the cross-sectional area of the ice block.

thus,

volume of the ice block, V = A(x + 25)

where,

x is the height of the ice block below the surface

Volume of the submerged ice block, V' = A × x

Total weight of the ice block, W = V ×   × g

where, g is the acceleration due to gravity

also,

Buoyant force acting on the ice block, = × V' × g

Now,

The weight of the block = buoyant force.

thus,

V ×   × g =   × V' × g

or

V ×    =   × V'

on substituting the values we get

A(x + 25)  ×  920  =  1025 × Ax

or

920x + 23000 = 1025x

x = 219.04 cm

Hence, the height of the ice block below the surface of water is 219.04 cm

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

well clustered together.

points close on a scatter diagram