/*! 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 47 A large aquarium of height \(5.0... [FREE SOLUTION] | 91Ó°ÊÓ

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A large aquarium of height \(5.00 \mathrm{~m}\) is filled with fresh water to a depth of \(2.00 \mathrm{~m}\). One wall of the aquarium consists of thick plastic \(9.00 \mathrm{~m}\) wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of \(4.00 \mathrm{~m}\) ?

Short Answer

Expert verified
The force increases by 529,020 N when the aquarium's water level rises from 2 m to 4 m.

Step by step solution

01

Identify the Problem

We need to find out how much the total force on one wall of the aquarium increases when the water depth changes from 2.00 m to 4.00 m.
02

Understand the Formula for Hydrostatic Force

The hydrostatic pressure at a certain depth in a fluid is given by the formula \( P = \rho g h \), where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid. Since the fluid is water, \( \rho = 1000 \) kg/m³ and \( g = 9.81 \) m/s².
03

Calculate the Initial Force at 2m Depth

The force on the wall from water is calculated as the integral of pressure over the area. For a wall of height \( h \) and width \( w \), the force is \( F = \rho g \int_{0}^{h} x \cdot w \cdot dx \). For a 2 m depth, calculate \( F_1 \).
04

Calculate the Final Force at 4m Depth

Similarly, calculate the force \( F_2 \) when the depth of the water is now 4 m by integrating \( P \) over the new depth.
05

Determine the Increase in Force

Calculate the increase in force as \( \Delta F = F_2 - F_1 \) using the outcomes of the previous two integration calculations.
06

Integrate to Find Forces

Perform the integration: \[F_1 = \int_{0}^{2} \rho g x \cdot 9.00 \; dx \F_2 = \int_{0}^{4} \rho g x \cdot 9.00 \; dx \]Both calculations yield the following:\[F_1 = 1000 \times 9.81 \times \frac{9}{2} \times \left(\frac{x^2}{2} \right)\bigg|_0^2 \]\[F_2 = 1000 \times 9.81 \times \frac{9}{2} \times \left(\frac{x^2}{2} \right)\bigg|_0^4 \]Calculate these values to get specific force values.
07

Perform Final Calculations

Substitute and compute for:- \( F_1: 1000 \times 9.81 \times 9 \times \left(\frac{4}{2}\right) = 176,580 \) N- \( F_2: 1000 \times 9.81 \times 9 \times \left(\frac{16}{2}\right)= 705,600 \) NThen calculate \( \Delta F = F_2 - F_1 = 705,600 - 176,580 = 529,020 \) N.

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Key Concepts

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

Understanding Pressure
Pressure is a concept that helps us understand how force is applied over a specific area. In the context of fluids like water, pressure depends on how deep the fluid is. This is because the weight of the water above pushes downwards, increasing the amount of force at greater depths. The formula for hydrostatic pressure is given by:
  • \( P = \rho g h \)
Here, \( P \) is the pressure, \( \rho \) is the density of the liquid (for water, it is 1000 kg/m³), \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( h \) is the depth of the liquid.
Using this formula, one can calculate how much pressure is exerted at a particular depth within the aquarium by simply multiplying these values together.
Thus, as the depth of the water increases, so does the pressure. This increase in pressure leads directly to an increase in the force applied on the walls of the aquarium.
Role of Integration in Calculating Force
Integration is a powerful mathematical tool used to calculate total values such as force when they change with respect to another variable, like depth.
In our problem, we use integration to find the total force on the aquarium wall at different water depths. The force exerted by the water at any particular point depends on the pressure, which changes with depth.
To calculate the total force (\( F \)) experienced by a wall, we set up an integral of pressure over the height of the wall:
  • \( F = \rho g \int_{0}^{h} x \cdot w \cdot dx \)
Here, \( x \) represents the depth variable, and \( w \) is the width of the wall.
For a change in the height of water from 2 m to 4 m, two separate integrals must be calculated. These integrals help determine the initial and new forces respectively.
This method shows how integration helps in calculating the total effect when there are varying amounts of pressure across a distance, like the height of an aquarium wall.
Physics Problem Solving Steps
When solving physics problems, especially those involving hydrostatic force, following a structured approach helps to clearly work through the calculation stages.
Here are key steps used in the problem:
  • Identify the core element of what needs to be found, such as the change in force when water depth changes in this case.
  • Understanding concepts and formulas, like hydrostatic pressure \( P = \rho g h \) and how they apply to the scenario.
  • Set up equations to represent the physical situation, such as integral equations for force.
  • Perform integration calculations to solve the equations. This involves functional steps like setting limits for depth and integrating pressure over the established range.
  • Finally, use basic arithmetic to find the increase in force using the results from your integrations.
By sequentially and peacefully working through these steps, complicated physics problems become manageable, and the principles become clearer.

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Most popular questions from this chapter

Giraffe bending to drink. In a giraffe with its head \(1.8 \mathrm{~m}\) above its heart, and its heart \(2.0 \mathrm{~m}\) above its feet, the (hydrostatic) gauge pressure in the blood at its heart is 250 torr. Assume that the giraffe stands upright and the blood density is \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).In torr (or \(\mathrm{mm} \mathrm{Hg}\) ), find the (gauge) blood pressure (a) at the brain (the pressure is enough to perfuse the brain with blood, to keep the giraffe from fainting) and (b) at the feet (the pressure must be countered by tight-fitting skin acting like a pressure stocking). (c) If the giraffe were to lower its head to drink from a pond without splaying its legs and moving slowly, what would be the increase in the blood pressure in the brain? (Such action would probably be lethal.)

The water flowing through a \(1.9 \mathrm{~cm}\) (inside diameter) pipe flows out through three \(1.5 \mathrm{~cm}\) pipes. (a) If the flow rates in the three smaller pipes are 26,21 , and \(16 \mathrm{~L} / \mathrm{min}\), 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 \(26 \mathrm{~L} / \mathrm{min}\) ?

A hollow sphere of inner radius \(8.0 \mathrm{~cm}\) and outer radius \(9.0 \mathrm{~cm}\) floats half-submerged in a liquid of density \(820 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.

How much work is done by pressure in forcing \(2.0 \mathrm{~m}^{3}\) of water through a pipe having an internal diameter of \(13 \mathrm{~mm}\) if the difference in pressure at the two ends of the pipe is \(1.0 \mathrm{~atm} ?\)

\(g-L O C\) in dogfights. When a pilot takes a tight turn at high speed in a modern fighter airplane, the blood pressure at the brain level decreases, blood no longer perfuses the brain, and the blood in the brain drains. If the heart maintains the (hydrostatic) gauge pressure in the aorta at 120 torr (or \(\mathrm{mm} \mathrm{Hg}\) ) when the pilot undergoes a horizontal centripetal acceleration of \(4.5 g\), what is the blood pressure (in torr) at the brain, \(30 \mathrm{~cm}\) radially inward from the heart? The perfusion in the brain is small enough that the vision switches to black and white and narrows to "tunnel vision" and the pilot can undergo \(\mathrm{g}\)-LOC (" \(g\)-induced loss of consciousness"). Blood density is \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).

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