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a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column is 16.0 cm high the oil air interfdace is 4.5 cm above the liquid level in the left arm of the u tube algrebraic expression to deter,ine the density of the unknown fluid

Sagot :

Answer:

The answer is "[tex]1155\ \frac{kg}{m^3}[/tex]"

Explanation:

Please find the complete question in the attached file.

[tex]p = p_0 + ?gh[/tex]

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

[tex]h_1 = 11 \ cm\\\\h_2= 3 \ cm[/tex]

The pressures would be proportional to the quantity [tex]11-3 = 8[/tex] cm from below the surface at the interface between both the oil and the liquid.

[tex]\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\[/tex]

       [tex]= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}[/tex]

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