The average density of the Earth.

Question

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Answer 1

Question

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.

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Answer 2

Question

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.

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Answer 3

Question

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.

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Answer 4

Question

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.

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Answer 5

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.

Answer 6

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

Answer 7

Chapter 1, Problem 1P

(a)

To determine

The average density of the Earth.

Answer 8

(a)

Expert Solution

Answer to Problem 1P

The average density of the Earth is $5.52×103 kg/m3$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

The average mass of the Earth is $5.97×1024 kg$ and the mean radius of the Earth is $6.37×106 m$ .

Formula to calculate the average density of the Earth is,

$ρ=mV$ (I)

$ρ$ is the average density of the Earth.
$m$ is the average mass of the Earth.
$V$ is the average volume of the Earth.

Formula to calculate the average volume of the Earth is,

$V=43πr3$

$r$ is the mean radius of the Earth.

Substitute $43πr3$ for $V$ in equation (I).

$ρ=m43πr3=3m4πr3$

Substitute $5.97×1024 kg$ for $m$ and $6.37×106 m$ for $r$ to find $ρ$ .

$ρ=3(5.97×1024 kg)4π(6.37×106 m)3=1.791×1025 kg3.24×1021 m3=5.52×103 kg/m3$

Conclusion:

Therefore, the average density of the Earth is $5.52×103 kg/m3$ .

Answer 13

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.

Conclusion:

Therefore, the average density of granite type rocks is less than the average density of the Earth.

Answer 14

(b)

To determine

The position of the density of Earth fit among those listed in Table 14.1 and compare the density of surface rocks like granite to the density of Earth.

Answer 15

(b)

Expert Solution

Answer to Problem 1P

The average density of Earth lies in between the densities of aluminum and iron and the average density of granite is less than the density of the Earth.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Section 1:

To find the position of the density of Earth fit among those listed in Table 14.1:

The average density of the Earth is $5.52×103 kg/m3$ . From the table 14.1, it lies in between the density of aluminum $2.70×103 kg/m3$ and the density of tin $7.30×103 kg/m3$ .

Conclusion:

Therefore, the average density of Earth lies in between the densities of aluminum and tin.

Section 2:

To determine the comparison of density of a typical surface rock like granite and the average density of Earth:

Introduction: The density of any substance is defined as the mass of the substance per unit volume.

The average density of the Earth is $5.52×103 kg/m3$ and the average density of the typical surface rocks like granite is $2.65×103 kg/m3$

Granite is a type rock that is mostly found on the surface of the earth.

Since the value $2.65×103 kg/m3$ is less than $5.52×103 kg/m3$ . Hence the density of granite type surface rocks is less than the average density of the Earth.