You come across an open container which is filled with two liquids. since the two liquids have different density there is a distinct separation between them. water fills the lower portion of the container to a depth of 0.209 m which has a density of 1.00 × 103 kg/m3. the fluid which is floating on top of the water is 0.300 m deep. if the absolute pressure on the bottom of the container is 1.049 × 105 pa, what is the density of the unknown fluid? the acceleration due to gravity is g = 9.81 m/s2 and atmospheric pressure is p0 = 1.013 × 105 pa.

526.57 Pa

Explanation:

P ( pressure at the bottom of the container) = 1.049 × 10^5 pa

Using the formula of pressure in an open liquid

Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²

Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa

P( atmospheric pressure) = 1.013 × 10^5 Pa

Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g

Subtract each of the pressure from the absolute pressure at the bottom

P(bottom) - atmospheric pressure

(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa

subtract pressure due to water from the remainder

3600 - 2050.29 = 1549.71 Pa

1549.71 =  ρ(density of the liquid) × h (depth of the liquid) × g

ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

where is the