/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 What gauge pressure must a machi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What gauge pressure must a machine produce in order to suck mud of density \(1800 \mathrm{~kg} / \mathrm{m}^{3}\) up a tube by a height of \(1.5 \mathrm{~m} ?\)

Short Answer

Expert verified
The gauge pressure required is approximately 26487 Pa.

Step by step solution

01

Identify the formula to use

To determine the gauge pressure needed to lift mud through a height, we will use the hydrostatic pressure formula: \[ P = \rho g h \]where \( P \) is the pressure, \( \rho \) is the density of the mud, \( g \) is the acceleration due to gravity, and \( h \) is the height the mud is lifted.
02

Assign known values to variables

We know the density of the mud \( \rho = 1800 \text{ kg/m}^3 \), the height \( h = 1.5 \text{ m} \), and the standard acceleration due to gravity \( g = 9.81 \text{ m/s}^2 \).
03

Calculate the gauge pressure

Substitute the known values into the hydrostatic pressure formula:\[ P = 1800 \times 9.81 \times 1.5 \]Now perform the multiplication:1. Calculate \( 1800 \times 1.5 = 2700 \).2. Then multiply by \( 9.81 \): \( 2700 \times 9.81 \approx 26487 \text{ Pa} \).
04

Finalize the result

Therefore, the gauge pressure required is approximately \( 26487 \text{ Pa} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure
Hydrostatic pressure is an essential concept in fluid mechanics, often applied when dealing with liquids at rest. This type of pressure is the force exerted by a fluid due to its weight. It increases with depth within a fluid because of the gravitational pull acting on the liquid above. Hydrostatic pressure is crucial when calculating the force needed to move or lift a fluid, like mud in our exercise.

Hydrostatic pressure depends on three main factors:
  • The density of the fluid (\(\rho\)
  • The height (\(h\)) or depth of the fluid
  • The force of gravity (\(g\))
This pressure is calculated using the formula: \[ P = \rho g h \] where each element in the formula is interlinked to determine the total pressure exerted by the fluid. Understanding this concept helps us comprehend why deeper fluids exert more pressure and why containers holding liquids are designed to withstand such forces.
Density
Density is a fundamental property of matter that describes how much mass a substance contains in a given volume. It is especially significant in calculating hydrostatic pressure, as it directly affects the amount of force a fluid can exert. In our exercise, the density of mud is given as \(1800 \, \text{kg/m}^3\), indicating it is quite dense compared to substances like water.

The formula for density is:\[ \rho = \frac{m}{V} \]where:
  • \(\rho\) is the density
  • \(m\) is the mass
  • \(V\) is the volume
In many practical situations, knowing the density helps in predicting whether a fluid can support or exert enough force to perform tasks like lifting mud through a tube. High-density fluids will exert more hydrostatic pressure for a given height and gravitational force, which is why understanding this value is vital in calculating pressure.
Pressure Formula
The pressure formula is an equation used to determine the pressure exerted by a fluid. In the context of hydrostatic conditions, it combines density, gravity, and height, giving us the specific calculation: \[ P = \rho g h \] This equation is powerful because it connects the physical properties of a fluid, like its density, to its capacity to exert a force over an area.

Breaking down:
  • \(P\) represents pressure measured in Pascals (Pa).
  • \(\rho\) is the fluid's density, contributing directly to its weight.
  • \(g\) is the acceleration due to gravity, which is usually \(9.81 \text{ m/s}^2\) on Earth's surface.
  • \(h\) is the height or depth, showing how much vertical distance the fluid covers.
By using this formula, we can understand how each variable impacts the total pressure a fluid generates, further aiding in engineering and scientific calculations necessary for applications and problem-solving.
Acceleration due to Gravity
The acceleration due to gravity is a natural phenomenon that affects all objects on Earth by drawing them towards the planet's center. Generally taken as \(9.81 \text{ m/s}^2\), it is an essential part of the pressure formula for hydrostatic conditions, as it directly influences the weight of a fluid, hence affecting its pressure.

Understanding gravity's role:
  • It is constant, making it a reliable factor in calculations.
  • Influences the force a fluid can exert downwards.
  • Used universally in physics for calculating forces and motion.
In our example, gravity's constant value ensures calculations are standardized and consistent. Knowing how gravity operates allows engineers and scientists to design systems and structures, such as tubes or containers, expecting the usual downward force and its effect on fluid movement.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block of wood has a mass of \(3.67 \mathrm{~kg}\) and a density of \(600 \mathrm{~kg} / \mathrm{m}^{3}\). It is to be loaded with lead \(\left(1.14 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\right)\) so that it will float in water with \(0.900\) of its volume submerged. What mass of lead is needed if the lead is attached to (a) the top of the wood and (b) the bottom of the wood?

Suppose that you release a small ball from rest at a depth of \(0.600 \mathrm{~m}\) below the surface in a pool of water. If the density of the ball is \(0.300\) that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)

The water flowing through a \(1.9 \mathrm{~cm}\) (inside diameter) pipe flows out through three \(1.3 \mathrm{~cm}\) pipes. (a) If the flow rates in the three smaller pipes are 26,19, and \(11 \mathrm{~L} / \mathrm{min}\), what is the flow rate in the \(1.9 \mathrm{~cm}\) pipe? (b) What is the ratio of the speed in the \(1.9 \mathrm{~cm}\) pipe to that in the pipe carrying \(26 \mathrm{~L} / \mathrm{min} ?\)

An iron anchor of density \(7870 \mathrm{~kg} / \mathrm{m}^{3}\) appears \(200 \mathrm{~N}\) lighter in water than in air. (a) What is the volume of the anchor? (b) How much does it weigh in air?

A garden hose with an internal diameter of \(1.9 \mathrm{~cm}\) is connected to a (stationary) lawn sprinkler that consists merely of a container with 24 holes, each \(0.13 \mathrm{~cm}\) in diameter. If the water in the hose has a speed of \(0.91 \mathrm{~m} / \mathrm{s}\), at what speed does it leave the sprinkler holes?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.