/*! 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 24 The piston of a hydraulic automo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 24

The piston of a hydraulic automobile lift is 0.30 m in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 kg? Also express this pressure in atmospheres.

Short Answer

Expert verified
166,352 Pa or 1.64 atm.

Step by step solution

01

Calculate the Force Required

First, we need to calculate the force required to lift the car. The formula for weight (force due to gravity) is given by \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity \( 9.8 \ m/s^2 \). For a car with a mass of 1200 kg: \[ F = 1200 \ kg \times 9.8 \ m/s^2 = 11,760 \ N \].
02

Calculate the Area of the Piston

The area \( A \) of the circular piston can be calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius. Given the diameter of the piston is 0.30 m, the radius \( r \) is 0.15 m. Therefore, \[ A = \pi \times (0.15 \ m)^2 \approx 0.0707 \ m^2 \].
03

Calculate Gauge Pressure in Pascals

Gauge pressure \( P \) is calculated using the formula \( P = \frac{F}{A} \). Using the force from Step 1 and the area from Step 2: \[ P = \frac{11,760 \ N}{0.0707 \ m^2} \approx 166,352 \ Pa \].
04

Convert Pressure to Atmospheres

To convert the pressure from pascals to atmospheres, use the conversion factor 1 atmosphere = 101,325 pascals. Therefore, \[ P = \frac{166,352 \ Pa}{101,325 \ Pa/atm} \approx 1.64 \ atm \].

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 91影视!

Key Concepts

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

Gauge Pressure
When lifting a car using a hydraulic lift, understanding gauge pressure is essential. Gauge pressure refers to the pressure relative to atmospheric pressure, which is typically the pressure we measure for practical purposes like this. Imagine that atmospheric pressure is a baseline level of pressure all around us. Whenever you exert pressure beyond this ambient level, we call it gauge pressure.
In mechanics, such as with the hydraulic lift example, it鈥檚 useful to talk about gauge pressure because it helps us understand the actual pressure exerting force on the lifting mechanism. The formula used for gauge pressure is simply the force applied divided by the area over which it is applied, expressed as: \( P = \frac{F}{A} \).
This calculation excludes atmospheric pressure, offering a clearer view of the specific force exertion scenario.
Force Calculation
Calculating force is a fundamental part of physics problems like the hydraulic lift. It helps to determine how much power is necessary to perform a task 鈥� in this case, lifting a car. Force is typically measured in newtons, derived from the mass of an object and the acceleration due to gravity. The formula to use here is \( F = mg \), where \( m \) represents mass and \( g \) is the acceleration due to gravity.
In this problem, with a car mass of 1200 kg, the gravitational acceleration is usually \( 9.8 \, m/s^2 \). Therefore, the force calculated comes out to \( 11,760 \, N \).
This calculation is pivotal because it sets the groundwork for understanding how much pressure needs to be applied to liftheavy objects effectively using a hydraulic system.
Pressure Conversion
Once you have your gauge pressure calculated in pascals, it's practical to express this pressure in more familiar units, such as atmospheres, for broader understanding. The conversion from pascals to atmospheres is straightforward with a known conversion factor 鈥� 1 atmosphere equals 101,325 pascals.
If you have calculated the gauge pressure in pascals, you can adjust it to atmospheres using: \( P_{atm} = \frac{P_{Pa}}{101,325} \).
For example, a gauge pressure of 166,352 pascals equates to approximately 1.64 atmospheres. By converting pressure units, you open up the information to a wider audience who might be more familiar with other units, which is particularly useful in fields like meteorology and oceanography. Recognizing the conversion process ensures precise and versatile communication of scientific data.

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

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform\(-\)so the hole is 50.0 cm above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \({p_g}\) of the air in the tank. (a) Graph \({R^2}\) as a function of \({p_g}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in kg/m\(^3\)). Use \(g\) \(=\) 9.80 m/s\(^2\). Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in h during the measurements is very small.

In a lecture demonstration, a professor pulls apart two hemispherical steel shells (diameter \(D\)) with ease using their attached handles. She then places them together, pumps out the air to an absolute pressure of \(p\), and hands them to a bodybuilder in the back row to pull apart. (a) If atmospheric pressure is \({p_0}\) , how much force must the bodybuilder exert on each shell? (b) Evaluate your answer for the case \(p\) \(=\) 0.025 atm, \(D\) \(=\) 10.0 cm.

A rock with density 1200 kg/m\(^3\) is suspended from the lower end of a light string. When the rock is in air, the tension in the string is 28.0 N. What is the tension in the string when the rock is totally immersed in a liquid with density 750 kg/m\(^3\)?

A cubical block of density \({\rho_B}\) and with sides of length \(L\) floats in a liquid of greater density \({\rho_L}\). (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \({\rho_W}\)) and does not mix with it. If water is poured on the surface of that liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L\), \({\rho_B}\), \({\rho_L}\), and \({\rho_W}\). (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and \(L\) \(=\) 10.0 cm.

(a) As you can tell by watching them in an aquarium, fish are able to remain at any depth in water with no effort. What does this ability tell you about their density? (b) Fish are able to inflate themselves using a sac (called the \(swim\) \(bladder\)) located under their spinal column. These sacs can be filled with an oxygen\(-\)nitrogen mixture that comes from the blood. If a 2.75-kg fish in freshwater inflates itself and increases its volume by 10%, find the \(net\) force that the \(water\) exerts on it. (c) What is the net \(external\) force on it? Does the fish go up or down when it inflates itself?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Mechanics and Materials

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Geometrical and Physical Optics

Read Explanation

Solid State Physics

Read Explanation
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.