The specific gravity of ice is 0.9. The area of the smallest slab of ice of height 0.5 m floating in fresh water that will just support a 100 kg man is
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9, which means the density of ice is 0.9 times the density of water. Show more…
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The specific gravity of ice is $0.9$. The area of the smallest slab of ice of height $0.5 \mathrm{~m}$ floating in fresh water that will just support a $100 \mathrm{~kg}$ man is (A) $1.5 \mathrm{~m}^{2}$ (B) $2 \mathrm{~m}^{2}$ (C) $3 \mathrm{~m}^{2}$ (D) $4 \mathrm{~m}^{2}$
A block of ice of area $A$ and thickness $0.5 \mathrm{~m}$ is floating in the fresh water. In order to just support a man of 100 $\mathrm{kg}$, the area $A$ should be : (the specific gravity of ice is $0.917$ and density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$ ) (a) $1.24 \mathrm{~m}^{2}$ (b) $4.21 \mathrm{~m}^{2}$ (c) $2.41 \mathrm{~m}^{2}$ (d) $7.23 \mathrm{~m}^{2}$
What is the minimum area (in square meters) of the top surface of an ice slab 0.441 m thick floating on fresh water that will hold up a 938 kg automobile? Take the densities of ice and fresh water to be 917 $\mathrm{kg} / \mathrm{m}^{3}$ and $998 \mathrm{kg} / \mathrm{m}^{3},$ respectively.
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University Physics with Modern Physics
Physics: Principles with Applications
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