Question

The specific gravity of ice is 0.9. Find the area(in m^2) of the smallest slab of ice of height 0.5m floating in freshwater, that will just support a 100kg man. plzz explain.

Answer 1

Answer:

Let the area be a . So volume of ice will be

0.5 × a = a/2 m3 .

Now calculate mass of this ice cube .

$specific \: \: gravity \: = \frac{ \rho(ice) }{ \rho(water)} \\ so\\ \rho(ice) = 1000 \times 0.9 = 900$

So mass = 900 × a/2 = 450a kg

Now see diagram buoyancy force must balance weight of ice and man .

$F_b = \rho \: Vg = 1000( \frac{a}{2} )10 = (100 + 450a)10\\500a=100+50a\\a=2m^2$

Answer 2

Explanation:

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