Fluids Physics Question: Calculating Density and Pressure of a Physics Book"

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Homework Help Overview

The discussion revolves around calculating the density and pressure exerted by a physics book based on its dimensions and weight. The problem involves applying principles from fluid mechanics and physics related to pressure and density.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for density and pressure, questioning the use of specific dimensions in the pressure calculation when the book is balanced on its spine. There are attempts to verify the correctness of the original poster's calculations.

Discussion Status

Some participants have provided feedback on the calculations, suggesting that the original poster may have used incorrect dimensions for the pressure calculation when the book is balanced. There is a focus on ensuring the right approach is taken, but no consensus has been reached regarding the correctness of the answers.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on checking assumptions related to the dimensions used in calculations.

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Homework Statement


A physics book has a height of 26.0 cm, a width of 22.4 cm, and a thickness of 3.10 cm.

(a) What is the density of the physics book if it weighs 18 N?
? kg/m3
(b) Find the pressure that the physics book exerts on a desktop when the book lies face up.
? Pa
(c) Find the pressure that the physics book exerts on the surface of a desktop when the book is balanced on its spine.
? Pa


Homework Equations



Divide the weight by gravity to get the mass.
18N/9.81 = 1.83 kg.
P = F/A

The Attempt at a Solution



Convert cm to meters:
(26.0cm)/100 = 0.26
22.4/100 = 0.224
3.10/100 = 0.031
Now divide
1.83/(0.26*0.224*0.031) = 1.83/0.00180544 = 1014 kg/m^3
P = F/A
area of the book = (0.224*0.26) = 0.5824 m^2
18/0.5824 = 30.9 Pa

When balanced on its spine, its width and thickness make up the area
(0.224*0.0310) = 0.006944
18/0.006944 = 2592 Pa
 
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You got the first two right, but I would say that you should use height in the last problem instead of width.
 
Like i said, I believe my answer is wrong. If someone would double check it.
 
That's what I'm saying. Your answer is wrong because you used the wrong number in your calculations. All you need to do is use the right one.
 

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