PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 9, Problem 87P

(a)

To determine

The scale readings in grams.

(a)

Expert Solution
Check Mark

Answer to Problem 87P

The scale reading is 81.1g.

Explanation of Solution

Dimensions of aluminum block is 2.00cm by 3.00cm by 5.00cm.

The spring scale shows the mass of block.

Write the equation to find the mass of block.

m=ρalV

Here, m is the mass of block, ρ is the density of aluminum, and V is the volume of block.

Conclusion:

Substitute 2702kg/m3 for ρal and 2.00cm×3.00cm×5.00cm for V in the above equation to find m.

m=(2702kg/m3)((2.00cm×3.00cm×5.00cm)(106m31cm3))=8.11×102kg(103g1kg)=81.1g

Therefore, the scale reading is 81.1g.

(b)

To determine

The readings of two scales.

(b)

Expert Solution
Check Mark

Answer to Problem 87P

Readings are 55.6g and 486g respectively.

Explanation of Solution

Density of oil is 850kg/m3 and reading of the second scale is 460.0g.

Write the equation to find the reading of first scale.

m1=mρoilV (I)

Here, m1 is the mass of first scale and ρoil is the density of oil.

Write the equation to find the scale reading with block immersed in the oil.

m2=mb+ρoilV (II)

Here, m2 is the scale reading with block immersed in the oil and mb is the mass of beaker with oil.

Conclusion:

Substitute 81.1g for m, 850kg/m3 for ρoil, and 2.00cm×3.00cm×5.00cm for V in equation (I) to find m1.

m1=81.1g((850kg/m3(103g1kg))((2.00cm×3.00cm×5.00cm)(106m31cm3)))=81.1g25.5g=55.6g

Substitute 460.0g for mb, 850kg/m3 for ρoil, and 2.00cm×3.00cm×5.00cm for V in equation (II) to find m2.

m2=460.0g+((850kg/m3)((2.00cm×3.00cm×5.00cm)(106m31cm3)))=460.0g+26.0g=486.0g

Therefore, the readings are 55.6g and 486g respectively.

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Chapter 9 Solutions

PHYSICS

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