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3 years ago

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A woman floats in a region of the Great Salt Lake where the water is about four times saltier than the ocean and has a density o

f about 1130 kg/m3. The woman has a mass of 52 kg, and her density is 943 kg/m3 after exhaling as much air as possible from her lungs.
1) Determine the percentage of her volume that will be above the waterline of the Great Salt Lake.

1 answer:

julsineya [31]3 years ago

3 0

To develop the problem it is necessary to use the concepts related to density and its definition with respect to mass and volume, as well

$\rho= \frac{m}{V}$

Therefore to calculate the volume of woman:

$V = \frac{m}{\rho}\\V = \frac{52}{943}\\V = 0.0551 m^3$

Now, volume of water of Great Salt Lake with the mass of 52 kg,

$V_i = \frac{52}{1130} \\V_i = 0.0460 m^3$

So, percentage of her volume changed will be,

$\%=\frac{(V - Vi)}{V}*100\\\% = \frac{(0.0551 - 0.0460)}{0.0551}*100 \\\% == 16.51 \%\\$

Therefore the percentage of her volume that will be above the waterline of the Great Salt Lake is 16.51%

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See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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