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never [62]
3 years ago
8

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.
Physics
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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Answer:

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Explanation:

given,

r = 539 m

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N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

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         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

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<h2>Answer: higher mean annual rainfall and temperatures. </h2>

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