EXAMPLE 9.5 Oil and Water DAL Calculate pressures created by layers of different fluids. Air ROBLEM In a huge oil tanker, salt water has flooded an oil tank to a depth h, = 5.00 m. On top of the water is a layer of oil h, = 8.00 m deep, as in e cross-sectional view of the tank in the figure. The oil has a density of 700 g/cm³. Find the pressure at the bottom of the tank. (Take 1,025 kg/m³ Po Oil P the density of salt water.) Water Rot TRATEGY P = Po+pgh must be used twice. First, use it to calculate the essure P, at the bottom of the oil layer. Then use this pressure in place of o in the equation and calculate the pressure Pbot at the bottom of the water layer. OLUTION (1) P, = Po+ pgh, = 1.01 x 105 Pa • (700 - 10° kg/m?) (9 80 m/s?)(8.00m) P, = 1.56 x 105 Pa se the equation to calculate the pressure at e bottom of the oil layer. 9.80 ow adapt the equation to the new starting essure, and use it to calculate the pressure (2) Pbot = P, + pgh2 = 1.56 x 105 Pa the bottom of the water layer. *(1025 - 10° kg/m³) (9.80 m/s?)(5.00 m) Pbot = 2.06 x 105 Pa EARN MORE EMARKS The weight of the atmosphere results in Po at the surface of the oil layer. Then the weight of the oil and the eight of the water combine to create the pressure at the bottom. UESTION Why does air pressure decrease with increasing altitude? (Select all that apply.) The weight of the air below is smaller.
EXAMPLE 9.5 Oil and Water DAL Calculate pressures created by layers of different fluids. Air ROBLEM In a huge oil tanker, salt water has flooded an oil tank to a depth h, = 5.00 m. On top of the water is a layer of oil h, = 8.00 m deep, as in e cross-sectional view of the tank in the figure. The oil has a density of 700 g/cm³. Find the pressure at the bottom of the tank. (Take 1,025 kg/m³ Po Oil P the density of salt water.) Water Rot TRATEGY P = Po+pgh must be used twice. First, use it to calculate the essure P, at the bottom of the oil layer. Then use this pressure in place of o in the equation and calculate the pressure Pbot at the bottom of the water layer. OLUTION (1) P, = Po+ pgh, = 1.01 x 105 Pa • (700 - 10° kg/m?) (9 80 m/s?)(8.00m) P, = 1.56 x 105 Pa se the equation to calculate the pressure at e bottom of the oil layer. 9.80 ow adapt the equation to the new starting essure, and use it to calculate the pressure (2) Pbot = P, + pgh2 = 1.56 x 105 Pa the bottom of the water layer. *(1025 - 10° kg/m³) (9.80 m/s?)(5.00 m) Pbot = 2.06 x 105 Pa EARN MORE EMARKS The weight of the atmosphere results in Po at the surface of the oil layer. Then the weight of the oil and the eight of the water combine to create the pressure at the bottom. UESTION Why does air pressure decrease with increasing altitude? (Select all that apply.) The weight of the air below is smaller.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Fluid Pressure
The term fluid pressure is coined as, the measurement of the force per unit area of a given surface of a closed container. It is a branch of physics that helps to study the properties of fluid under various conditions of force.
Gauge Pressure
Pressure is the physical force acting per unit area on a body; the applied force is perpendicular to the surface of the object per unit area. The air around us at sea level exerts a pressure (atmospheric pressure) of about 14.7 psi but this doesn’t seem to bother anyone as the bodily fluids are constantly pushing outwards with the same force but if one swims down into the ocean a few feet below the surface one can notice the difference, there is increased pressure on the eardrum, this is due to an increase in hydrostatic pressure.
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Question
Transcribed Image Text:EXAMPLE 9.5
Oil and Water
GOAL Calculate pressures created by layers of different fluids.
Air
PROBLEM In a huge oil tanker, salt water has flooded an oil tank to a depth
of h, = 5.00 m. On top of the water is a layer of oil h, = 8.00 m deep, as in
Po
Oil
the cross-sectional view of the tank in the figure. The oil has a density of
P
0.700 g/cm3. Find the pressure at the bottom of the tank. (Take 1,025 kg/m3
as the density of salt water.)
Water
Rot
STRATEGY P = P,+pgh must be used twice. First, use it to calculate the
pressure P, at the bottom of the oil layer. Then use this pressure in place of
P in the equation and calculate the pressure Phot at the bottom of the water layer.
bot
SOLUTION
Use the equation to calculate the pressure at
(1) P, = Po+pgh,
the bottom of the oil layer.
= 1.01 x 105 Pa
102 kg/m3
9.80 m/s
P, = 1.56 x 105 Pa
Now adapt the equation to the new starting
pressure, and use it to calculate the pressure
(2) Pbot = P, + pgh2
= 1.56 x 105 Pa
at the bottom of the water layer.
*(1.025 « 10° kg/m³) (9.80 m/s²
m/s?)(s00m)
Pbot = 2.06 x 105 Pa
LEARN MORE
REMARKS The weight of the atmosphere results in P at the surface of the oil layer. Then the weight of the oil and the
weight of the water combine to create the pressure at the bottom.
QUESTION Why does air pressure decrease with increasing altitude? (Select all that apply.)
The weight of the air below is smaller.
The weight of the air above is smaller.
O Less air above pushes down.
O There is more air below to push up more strongly.
O Colder air is less dense.
O Hot air rises.
PRACTICE IT
Use the worked example above to help you solve this problem. In a huge oil tanker, salt water has flooded an oil tank to a
depth of h, = 4.50 m. On top of the water is a layer of oil h, = 8.25 m deep, as in the cross-sectional view of the tank as
shown in the figure. The oil has a density of 0.700 g/cm3. Find the pressure at the bottom of the tank. (Take 1,025 kg/m3
as the density of salt water.)
1.01e+05
Your response differs from the correct answer by more than 10%. Double check your calculations. Pa
EXERCISE
HINTS: GETTING STARTED | 'M STUCK!
Calculate the pressure on the top lid of a chest buried under 3.85 meters of mud with density 1.75 x 103 kg/m3 at the
bottom of a 12.5-m-deep lake.
Pa
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