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
Q&A
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?
Express your answer with the appropriate units.
1 Answer
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
get an exact answer.
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
Textbook: Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Edition: 3
Author: Randall D. Knight
ISBN: 9780321740908
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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
A 2.0-cm times 2.0-cm times 6.0-cm stick floats in water with its long axis vertical. The length...
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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.
Answer and Explanation:
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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