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# a 2.0cm×2.0cm×9.0cm block floats in water with its long axis vertical. the length of the block above water is 1.0 cm .

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## A 2.0cm×2.0cm×9.0cm block floats in water with its long axis vertical. The length of the block above water is?

A 2.0cm×2.0cm×9.0cm block floats in water with its long axis vertical. The length of the block above water is? - A 2.0cm×2.0cm×9.0cm block floats in water

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## A 2.0cm×2.0cm×9.0cm block floats in water with its long axis vertical. The length of the block above water is?

A 2.0cm×2.0cm×9.0cm block floats in water with its long axis vertical. The length of the block above water is

Part A

What is the block’s mass density?

The density of the block is the ratio of volume submerged

divided by the total volume (which will be less than 1gram/cm³)

Simple as that.

If you provide the length of the block above the water you can

Example: 2cm×2cm×9cm = 36cm³

If 1cm is exposed, then the volume submerged is

2cm×2cm×8cm × 32cm³ 32cm³/36cm³ = 8/9

The density of the block is 8/9×density of water.

Water density = 1g/cm³

Block density = 8/9g/cm³

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## Section 15.4 BuoyancyA 2.0 cm × 2.0 cm × 6.0 cm block

Section 15.4 BuoyancyA 2.0 cm × 2.0 cm × 6.0 cm block floats in water with its long axis vertical. The length of the block above water is 2.0 cm. What is the block’s mass density? Step 1 of 2We have to find the block’s mass density, if a 2.0 cm × 2.0 cm × 6.0 cm block floats in water with its long axis vertical and

Textbooks / Physics / Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 3 / Chapter 15 / Problem 15E

## Section 15.4 BuoyancyA 2.0 cm × 2.0 cm × 6.0 cm block

ISBN: 9780321740908 69

## Solution for problem 15E Chapter 15

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics | 3rd Edition

Get Full Solutions 318 Reviews 27 0 Problem 15E Problem 15E

Section 15.4 Buoyancy

A 2.0 cm × 2.0 cm × 6.0 cm block floats in water with its long axis vertical. The length of the block above water is 2.0 cm. What is the block’s mass density?

Step-by-Step Solution:

Step 1 of 2

We have to find the block’s mass density, if a 2.0 cm × 2.0 cm × 6.0 cm block floats in water with its long axis vertical and the length of the block above water is 2.0 cm.

The buoyant force on the cylinder is given by Archimedes principle. The cylinder is in static equilibrium so that the buoyant force is the equal to the weight of the displaced water.

FB=FG Where, FB= Buoyant force =ρwVwg FG=

weight of the displaced water

=mg=ρcylVcylg ρw= density of water =1000 kg/m3 Vw=

Volume of water displaced

in m3 ρcy=

density of block's mass in

kg/m3 Vcyl=

Volume of immersed block in

m3 g=9.80 m/s2 Step 2 of 2

### Chapter 15, Problem 15E is Solved

View Full Solution

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## A 2.0

Answer to: A 2.0-cm times 2.0-cm times 6.0-cm stick floats in water with its long axis vertical. The length of the stick above the water is 2.0-cm....

Density

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## Question:

A 2.0 − c m × 2.0 − c m × 6.0 − c m

2.0−cm×2.0−cm×6.0−cm

stick floats in water with its long axis vertical. The length of the stick above the water is

2.0 − c m 2.0−cm

. What is the stick's mass density?

## Mass Density:

In fluid mechanics, the mass density of the object can be defined as the fraction of the mass of the object to the volume of the object. Typically, the standard measurement unit for the mass density is kilogram per cubic centimeter.

The length of the stick above the water is

l = 2.0 c m l=2.0cm .

The length of the stick below the water is,

L = H − l = 6.0 − 2.0 = 4.0 c m L=H−l=6.0−2.0=4.0cm

The expression for the volume of the stick below water is,

V = 2.0 c m × 2 .0 c m × 4 .0 c m = 16 c m 3

V=2.0cm×2.0cm×4.0cm=16cm3

The expression for the amount of water displaced is,

m = V × ρ w = 16 c m 3 × 1 g / c m 3 = 16 g

m=V×ρw=16cm3×1g/cm3=16g

The mass of the block is equal to the mass of the water displaced.

m b = m = 16 g mb=m=16g

The volume of the stick floats in water is,

V b = 2.0 c m × 2 .0 c m × 6 .0 c m = 24 .0 c m 3

Vb=2.0cm×2.0cm×6.0cm=24.0cm3

The expression for the stick's mass density is,

ρ = m b V b ρ=mbVb

Substituting the given values in the above expression, we will get

ρ = 16 g 24 c m 3 = 0.667 g / c m 3 ≈ 0.67 g / c m 3

ρ=16g24cm3=0.667g/cm3≈0.67g/cm3

Thus, the stick's mass density is

0.67 g / c m 3 0.67g/cm3 .

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Learn about the density of gases. Explore and reorganize the ideal gas law formula to find gas density formula, and understand how to find the density of gases.

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James 12 month ago

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