PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 9, Problem 37P

(a)

To determine

The specific gravity of the disk.

(a)

Expert Solution
Check Mark

Answer to Problem 37P

The specific gravity of the disk is 0.910.

Explanation of Solution

The specific gravity of a substance is the ratio of its density to the density of water.

Write the equation for specific gravity of the disk.

S.G=ρρw (I)

Here, ρ is the density of the object and ρw is the density of water.

Write the equation for density.

ρ=mV (II)

Here, m is the mass of the disk and V is the volume of the cylindrical disk.

Conclusion:

Substitute equation (II) in equation (I).

S.G=mρwV. (III)

Substitute 8.16kg for m, 1.00×103kg/m3 for ρw, and 8.97×103m3 for V in equation (III).

S.G=8.16kg(1.00×103kg/m3)(8.97×103m3)=8.16kg8.97kg=0.910

Therefore, the specific gravity of the disk is 0.910.

(b)

To determine

The distance of the water level at its bottom surface.

(b)

Expert Solution
Check Mark

Answer to Problem 37P

The distance of the water level at its bottom surface is 1.28cm.

Explanation of Solution

Write the equation for buoyant force is equal to the weight of the fluid displaced.

F=0ρfgViρogVo=0 (IV)

Here, ρf is the density of the water at 0°C, g is the gravitational acceleration, Vi is the volume of the fluid displaced after immersed, ρo is the density of the object, and Vo is the volume of the object.

Since the relationship between the fraction of a floating object’s volume that is submerged and the ratio of the object’s density to that of the fluid in which it floats is rewritten from the equation (IV).

ρfgVf=ρogVoρfVf=ρoVoVfVo=ρoρf (V)

The ratio of the density of the disk to that of water is equal to the fraction of the volume of the disk is submerged.

VwaterVdisk=ρdiskρwater (VI)

Here, ρdisk is the density of the disk, Vdisk is the volume of the disk, ρwater is the density of the water, and Vwater is the volume of the water.

Compare equation (VI) and (I).

S.G=VwaterVdisk

Replace dA for Vwater and V for  Vdisk in the above equation.

S.G=dAV (VII)

Here, d is the distance of the water level at its bottom surface.

Solve the equation (VII) for d.

S.G=dAVd=(S.G)VA (VIII)

Conclusion:

Substitute 0.640m2 for A, 0.910 for S.G, and 8.97×103m3 for V in equation (VIII).

d=(0.910)(8.97×103m3)(0.640m2)=(0.910)(14.02×103m)=12.76×103m

Convert the above value into centimeter.

d=1.28×102m=1.28×102m(102cm1m)=1.28cm

Therefore, the distance of the water level at its bottom surface is 1.28cm.

(c)

To determine

The distance of the water level at its top surface.

(c)

Expert Solution
Check Mark

Answer to Problem 37P

The distance of the water level at its top surface is 0.13cm.

Explanation of Solution

From the equation (VII), find the distance of the top surface of the disk above the water level.

S.G=dAV

Replace hA for V in the above equation.

S.G=dAhA=dh (IX)

Here, h is the thickness of the disk

Rewrite equation (IX) by using equation (VIII).

hd=h(S.G)hh=VA(1S.G) (X)

Conclusion:

Substitute 0.640m2 for A, 0.910 for S.G, and 8.97×103m3 for V in equation (X).

h=8.97×103m30.640m2(10.910)=14.02×103m(0.09)=1.26×103m

Convert the above value into centimeter.

h=0.126×102m=0.13×102m(102cm1m)=0.13cm

Therefore, the distance of the water level at its top surface is 0.13cm.

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