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2 years ago

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10 pts A cube of water ice (rho = 0.917 × 103 kg/m3) is placed in mercury (rho = 13.6 × 103 kg/m3), which is liquid at 0° Celsiu

s. If we ignore any possible melting of the ice cube and problems with the surface tension of mercury, the fraction of the ice cube that floats above the surface of the mercury is

1 answer:

Margarita [4]2 years ago

8 0

Answer:

fraction of ice cube float is 0.933

Explanation:

given data

density of ice = 0.917 × 10³ kg/m³

density of mercury = 13.6 × 10³ kg/m³

to find out

what is fraction of the ice cube floats

solution

we find first ice submerged volume that is

ice submerged volume = $\frac{density of ice}{density of mercury}$

put here value

ice submerged volume = $\frac{0.917*10^{3} }{13.6*10^{3} }$

ice submerged volume = 0.0674

so here fraction of ice cube float is

fraction = 1 - 0.0674

fraction = 0.933

so fraction of ice cube float is 0.933

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where R is negative since its direction is opposite to the motion

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Learn more about Newton's second law:

brainly.com/question/3820012

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