The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

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Answer 1

Question

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

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Answer 2

Question

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

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Answer 3

Question

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

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Answer 4

Question

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Answer 6

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Answer 7

Chapter 1, Problem 1.25P

To determine

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Answer 8

Expert Solution & Answer

Answer to Problem 1.25P

The radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The mass of one cubic meter aluminum is $2.70×103 kg$ and the mass of one cubic meter iron is $7.86×103 kg$ . The radius of the solid iron sphere is $2.0 cm$ .

Write the formula to calculate the density of aluminum sphere

$ρalu=m1V1$

Here, $ρalu$ is the density of the aluminum sphere, $m1$ is the mass of one cubic meter aluminum sphere and $V1$ is the volume of the aluminum sphere.

Substitute $2.70×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρalu$ .

$ρalu=2.70×103 kg1.0 m3=2.70×103 kg/m3$

Write the formula to calculate the density of iron sphere

$ρiron=m2V2$

Here, $ρiron$ is the density of the iron sphere, $m2$ is the mass of one cubic meter iron sphere and $V2$ is the volume of the iron sphere.

Substitute $7.86×103 kg$ for $m1$ and $1.0 m3$ for $V1$ to find $ρiron$ .

$ρiron=7.86×103 kg1.0 m3=7.86×103 kg/m3$

Write the formula to calculate the mass of an aluminum sphere

$malu=ρalu⋅Valu$

Here, $malu$ is the mass of the aluminum sphere, $Valu$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of an aluminum sphere

$Valu=43πralu3$

Here, $ralu$ is the radius of the aluminum sphere.

Substitute $43πralu3$ for $Valu$ .

$malu=ρalu⋅43πralu3=43πralu3ρalu$ (I)

Write the formula to calculate the mass of a solid iron sphere of radius $2.0 cm$

$miron=ρiron⋅Viron$

Here, $miron$ is the mass of the aluminum sphere, $Viron$ is the volume of the aluminum sphere.

Write the formula to calculate the volume of iron sphere

$Viron=43πriron3$

Here, $riron$ is the radius of the iron sphere.

Substitute $43πriron3$ for $Viron$ .

$miron=ρiron⋅43πriron3=43πriron3ρiron$ (II)

Since both sphere must balance to each other on an equal arm balance. So, they both have equal mass.

Equating equation (I) and equation (II),

$malu=miron43πralu3ρalu=43πriron3ρironralu3ρalu=riron3ρironralu=ρironρalu3⋅riron$

Conclusion:

Substituting $2.0 cm$ for $riron$ , $7.86×103 kg/m3$ for $ρiron$ and $2.70×103 kg/m3$ for $ρalu$ in the above equation to find $ralu$ .

$ralu=(7.86×103 kg/m32.70×103 kg/m33)⋅(2.0 cm)=1.427×2.0 cm=2.86 cm$

Therefore, the radius of a solid aluminum sphere is $2.86 cm$ that balance a solid iron sphere on an equal arm balance.

Answer 12

Ch. 1 - In a machine shop, two cams are produced, one of...Ch. 1 - True or False: Dimensional analysis can give you...Ch. 1 - The distance between two cities is 100 mi. What is...Ch. 1 - One student uses a meterstick to measure the...Ch. 1 - A house is advertised as having 1 420 square feet...Ch. 1 - Answer each question yes or no. Must two...Ch. 1 - The price of gasoline at a particular station is...Ch. 1 - Rank the following five quantities in order from...Ch. 1 - Prob. 1.6OQ Ch. 1 - Prob. 1.7OQ

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Concept explainers

The radius of a solid aluminum sphere that balance a solid iron sphere on an equal arm balance.

Answer to Problem 1.25P

Explanation of Solution

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Chapter 1 Solutions