Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 1, Problem 25P
To determine

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

Expert Solution & Answer
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Answer to Problem 25P

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

Explanation of Solution

Section 1:

To determine: The density of the aluminum sphere.

Answer: The density of the aluminum sphere is 2.70×103kg/m3 .

Given information:

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

Formula to calculate the density of aluminum sphere is,

ρalu=m1V1

  • ρalu is the density of the aluminum sphere.
  • m1 is the mass of one cubic meter aluminum sphere.
  • V1 is the volume of the aluminum sphere.

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

ρalu=2.70×103kg1.0m3=2.70×103kg/m3

Section 2:

To determine: The density of the iron sphere.

Answer: The density of the iron sphere is 7.86×103kg/m3 .

Given information:

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

Formula to calculate the density of iron sphere is,

ρiron=m2V2

  • ρiron is the density of the iron sphere.
  • m2 is the mass of one cubic meter iron sphere.
  • V2 is the volume of the iron sphere.

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

ρiron=7.86×103kg1.0m3=7.86×103kg/m3

Section 3:

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

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

Given information:

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

Formula to calculate the mass of a aluminum sphere is,

malu=ρaluValu

Formula to calculate the volume of a aluminum sphere is,

Valu=43πralu3

  • ralu is the radius of the aluminum sphere.

Substitute 43πralu3 for Valu .

malu=ρalu43πralu3=43πralu3ρalu (I)

Formula to calculate the mass of a solid iron sphere of radius 2.0cm is,

miron=ρironViron

Formula to calculate the volume of iron sphere is,

Viron=43πriron3

  • riron is the radius of the iron sphere.

Substitute 43πriron3 for Viron .

miron=ρiron43πriron3=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=ρironρalu3riron

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

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

Conclusion:

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

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Chapter 1 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 1 - Prob. 8OQCh. 1 - Prob. 9OQCh. 1 - Prob. 10OQCh. 1 - Prob. 1CQCh. 1 - Prob. 2CQCh. 1 - Prob. 3CQCh. 1 - Prob. 4CQCh. 1 - Prob. 1PCh. 1 - Prob. 2PCh. 1 - Prob. 3PCh. 1 - Prob. 4PCh. 1 - Prob. 5PCh. 1 - Prob. 6PCh. 1 - A crystalline solid consists of atoms stacked up...Ch. 1 - Prob. 8PCh. 1 - Prob. 9PCh. 1 - Prob. 10PCh. 1 - Prob. 11PCh. 1 - Prob. 12PCh. 1 - The position of a particle moving under uniform...Ch. 1 - (a) Assume the equation x = At3 + Bt describes the...Ch. 1 - A solid piece of lead has a mass of 23.94 g and a...Ch. 1 - Prob. 16PCh. 1 - Prob. 17PCh. 1 - Prob. 18PCh. 1 - Why is the following situation impossible? A...Ch. 1 - Prob. 20PCh. 1 - Prob. 21PCh. 1 - Prob. 22PCh. 1 - Prob. 23PCh. 1 - Prob. 24PCh. 1 - Prob. 25PCh. 1 - Let AI represent the density of aluminum and Fe...Ch. 1 - One gallon of paint (volume = 3.78 103 m3) covers...Ch. 1 - Prob. 28PCh. 1 - Prob. 29PCh. 1 - Prob. 30PCh. 1 - Prob. 31PCh. 1 - (a) Compute the order of magnitude of the mass of...Ch. 1 - To an order of magnitude, how many piano tuners...Ch. 1 - Prob. 34PCh. 1 - Prob. 35PCh. 1 - How many significant figures are in the following...Ch. 1 - The tropical year, the time interval from one...Ch. 1 - Prob. 38PCh. 1 - Review. In a community college parking lot, the...Ch. 1 - Prob. 40PCh. 1 - Prob. 41PCh. 1 - Prob. 42PCh. 1 - Review. The ratio of the number of sparrows...Ch. 1 - Prob. 44PCh. 1 - Prob. 45PCh. 1 - Prob. 46PCh. 1 - Prob. 47PCh. 1 - Prob. 48PCh. 1 - Prob. 49PCh. 1 - Prob. 50PCh. 1 - Prob. 51PCh. 1 - Prob. 52PCh. 1 - Prob. 53PCh. 1 - Prob. 54APCh. 1 - Prob. 55APCh. 1 - Prob. 56APCh. 1 - Prob. 57APCh. 1 - Prob. 58APCh. 1 - Prob. 59APCh. 1 - Prob. 60APCh. 1 - Prob. 61APCh. 1 - The distance from the Sun to the nearest star is...Ch. 1 - Prob. 63APCh. 1 - A spherical shell has an outside radius of 2.60 cm...Ch. 1 - Prob. 65APCh. 1 - Air is blown into a spherical balloon so that,...Ch. 1 - Prob. 67APCh. 1 - In physics, it is important to use mathematical...Ch. 1 - The consumption of natural gas by a company...Ch. 1 - A woman wishing to know the height of a mountain...Ch. 1 - Prob. 71APCh. 1 - Prob. 72APCh. 1 - Prob. 73AP
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