The image presents a scenario with a horizontal pipe that has a varying cross section. A liquid with a density of \(1.65 \, \text{g/cm}^3\) flows from left to right, moving from a larger cross-sectional area to a smaller one.- **Left Side Properties:** - Cross-sectional area: \(10.0 \, \text{cm}^2\) - Speed of the liquid: \(2.46 \, \text{m/s}\) - Pressure: \(1.20 \times 10^5 \, \text{Pa}\)- **Right Side Properties:** - Cross-sectional area: \(3.00 \, \text{cm}^2\)The pipe illustration shows the flow of liquid through a constriction from a wider area to a narrower area, visualized with arrows indicating direction and speed.**Questions:**(a) What is the speed (in m/s) of the liquid in the right side (the smaller section)? (Enter your answer to at least three significant figures.)\[\boxed{\phantom{1}}\](b) What is the pressure (in Pa) of the liquid in the right side (the smaller section)?\[\boxed{\phantom{1}}\]

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Description: The image presents a scenario with a horizontal pipe that has a varying cross section. A liquid with a density of \(1.65 \, \text{g/cm}^3\) flows from left to right, moving from a larger cross-sectional area to a smaller one.- **Left Side Properties

Given: The density of liquid, ρ=1.65 gcm3=1.65 gcm3×10-3 kg1 g×1 cm3 10-6 m3=1650 kgm3 The cross-section area of left side, A1=10 cm2 The speed of liquid in left side, v1=2.46 ms The pressure at left side, P1=1.20×105 Pa The cross-section area of right side, A2=3 cm2 To determine: (a)The speed (in m/s) of the liquid in the right side. (b)The pressure (in Pa) of the liquid on the right side.Explanation: (a) Apply continuity equation at left and right side, we get A1v1=A2v2 Substitute the value in the above expression, the speed of liquid in the right side is calculated as: 10 cm2×2.46 ms=3 cm2×v2v2=24.63 =8.2 ms Thus, the speed of the liquid in the right side is 8.20 ms(b) The expression of Bernoulli equation between left and right side of pipe is written as: P1+12ρv12+ρgh1=P2+12ρv22+ρgh2 As pipe is lies along horizontal plane. So, h1=h2. The above expression now becomes, P1+12ρv12=P2+12ρv22 The pressure of the liquid on the right side is calculated as: 1.2×105 Pa+12×1650 kgm3×2.46 ms2=P2+12×1650 kgm3×8.2 ms2P2=1.2×105 Pa+12×1650 kgm3×6.0516-67.24P2=1.2×105 Pa-0.505×105 PaP2=0.695×105 Pa Thus, the pressure of the liquid on the right side is 0.695×105 Pa.

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The figure below shows a horizontal pipe with a varying cross section. A liquid with a density of 1.65 g/cm3 flows from left to right in the pipe, from larger to smaller cross section. The left side's cross-sectional area is 10.0 cm2, and while in this side, the speed of the liquid is 2.46 m/s, and the pressure is 1.20 x 105 Pa. The right side's cross sectional-area is 3.00 cm2. (a) What is the speed (in m/s) of the liquid in the right side (the smaller section)? (Enter your answer to at least three significant figures.) m/s (b) What is the pressure (in Pa) of the liquid in the right side (the smaller section)? Pa

The figure below shows a horizontal pipe with a varying cross section. A liquid with a density of 1.65 g/cm3 flows from left to right in the pipe, from larger to smaller cross section. The left side's cross-sectional area is 10.0 cm2, and while in this side, the speed of the liquid is 2.46 m/s, and the pressure is 1.20 x 105 Pa. The right side's cross sectional-area is 3.00 cm2. (a) What is the speed (in m/s) of the liquid in the right side (the smaller section)? (Enter your answer to at least three significant figures.) m/s (b) What is the pressure (in Pa) of the liquid in the right side (the smaller section)? Pa

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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Concept explainers

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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The image presents a scenario with a horizontal pipe that has a varying cross section. A liquid with a density of \(1.65 \, \text{g/cm}^3\) flows from left to right, moving from a larger cross-sectional area to a smaller one.

- **Left Side Properties:**
- Cross-sectional area: \(10.0 \, \text{cm}^2\)
- Speed of the liquid: \(2.46 \, \text{m/s}\)
- Pressure: \(1.20 \times 10^5 \, \text{Pa}\)

- **Right Side Properties:**
- Cross-sectional area: \(3.00 \, \text{cm}^2\)

The pipe illustration shows the flow of liquid through a constriction from a wider area to a narrower area, visualized with arrows indicating direction and speed.

**Questions:**

(a) What is the speed (in m/s) of the liquid in the right side (the smaller section)? (Enter your answer to at least three significant figures.)

\[
\boxed{\phantom{1}}
\]

(b) What is the pressure (in Pa) of the liquid in the right side (the smaller section)?

\[
\boxed{\phantom{1}}
\]

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