Physics for Scientists and Engineers, Volume 1
Physics for Scientists and Engineers, Volume 1
9th Edition
ISBN: 9781133954156
Author: Raymond A. Serway
Publisher: CENGAGE L
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Chapter 6, Problem 6.69CP

(a)

To determine

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to 10.0μm .

(a)

Expert Solution
Check Mark

Answer to Problem 6.69CP

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to 10.0μm is 0.0132m/sec .

Explanation of Solution

Given info: The expression of magnitude of resistive force exerted on a sphere is F=(arv+br2v2)newtons , the radius of the sphere is rmeters , the speed of the stream of air is vm/sec , the numerical value of constant a is 3.10×104 , the numerical value of constant b is 0.870 and the radii of the water droplets is 10.0μm .

The expression of the resistive force is,

F=(arv+br2v2)N

Here,

a is a constant.

b is a constant.

r is the radii of the droplet.

v is the terminal speed of the water droplet.

Substitute 3.10×104 for a and 0.870 for b in the above equation.

F=(3.10×104rv+0.870r2v2)N (1)

The expression of mass of the water droplet is,

m=ρV (2)

Here,

ρ is the density of the water.

V is the volume of the water droplet.

The expression of volume of the water droplet is,

V=43πr3

Substitute 10.0μm for r in the above equation.

V=43π[10.0μm(106m1μm)]3=43π(10.0×106m)3=4.188×1015m3V4.2×1015m3

Substitute 1.0×103kg/m3 for ρ and 4.2×1015m3 for V in equation (2).

m=(1.0×103kg/m3)(4.2×1015m3)m=4.2×1012kg

Thus, the mass of the water droplet is 4.2×1012kg .

The expression of force that acts on the water droplet is,

F=mg

Here,

g is the acceleration due to gravity.

Compare and equate the equation (1) and the above equation.

mg=(3.10×104rv+0.870r2v2)N (3)

Substitute 10.0μm for r , 4.2×1012kg for m and 9.8m/s2 for g in the above equation.

(4.2×1012kg)(9.8m/s2)=(3.10×104(10.0μm(106m1μm))v+0.870(10.0μm(106m1μm))2v2)N 41.16×1012=(3.10×104(10.0×106m)v+0.870(10.0×106m)2v2)41.16×1012=(3.10×109v+0.870×1010v2)

The contribution of the second term of v to the air resistance is very small and it is ignored. The new equation is,

41.16×1012=(3.10×109v)v=41.16×10123.10×109m/sec=13.277×103m/secv0.0132m/sec

Conclusion:

Therefore, the terminal speed for water droplets falling under their own weight in air for the drop radii equal to 10.0μm is 0.0132m/sec .

(b)

To determine

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to 100μm .

(b)

Expert Solution
Check Mark

Answer to Problem 6.69CP

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to 100μm is 1.03m/sec .

Explanation of Solution

Given info: The expression of magnitude of resistive force exerted on a sphere is F=(arv+br2v2)newtons , the radius of the sphere is rmeters , the speed of the stream of air is vm/sec , the numerical value of constant a is 3.10×104 , the numerical value of constant b is 0.870 and the radii of the water droplets is 100μm .

From equation (1) the expression of resistive force is,

F=(3.10×104rv+0.870r2v2)N

The expression of volume of the water droplet is,

V=43πr3

Substitute 100μm for r in the above equation.

V=43π[100μm(106m1μm)]3=43π(100×106m)3=4.188×1012m3V4.2×1012m3

From equation (2), the expression of mass of the water droplet is,

m=ρV

Substitute 1.0×103kg/m3 for ρ and 4.2×1012m3 for V in above equation.

m=(1.0×103kg/m3)(4.2×1012m3)m=4.2×109kg

From equation (3) the final equation is,

mg=(3.10×104rv+0.870r2v2)N

Substitute 100μm for r , 4.2×109kg for m and 9.8m/s2 for g in the above equation.

(4.2×109kg)(9.8m/s2)=(3.10×104(100μm(106m1μm))v+0.870(100μm(106m1μm))2v2)N 41.1×109=(3.10×104(100×106m)v+0.870(100×106m)2v2)4.11×108=(3.10×108v+0.870×108v2)

Further solve the above equation.

0.870×108v2+3.10×108v4.11×108=00.870v2+3.10v4.11=0

Apply quadratic formula to solve the above equation.

v=3.10±(3.10)24(0.870)(4.11)2(0.870)v1.03m/sec

Conclusion:

Therefore, the terminal speed for water droplets falling under their own weight in air for the drop radii equal to 100μm is 1.03m/sec .

(c)

To determine

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to 1.00mm .

(c)

Expert Solution
Check Mark

Answer to Problem 6.69CP

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to 1.00mm is 6.87m/sec .

Explanation of Solution

Given info: The expression of magnitude of resistive force exerted on a sphere is F=(arv+br2v2)newtons , the radius of the sphere is rmeters , the speed of the stream of air is vm/sec , the numerical value of constant a is 3.10×104 , the numerical value of constant b is 0.870 and the radii of the water droplets is 1.00mm .

From equation (1) the expression of resistive force is,

F=(3.10×104rv+0.870r2v2)N

The expression of volume of the water droplet is,

V=43πr3

Substitute 1.00mm for r in the above equation.

V=43π[1.00mm(103m1mm)]3=43π(1.00×103m)3=4.188×109m3V4.2×109m3

From equation (2), the expression of mass of the water droplet is,

m=ρV

Substitute 1.0×103kg/m3 for ρ and 4.2×109m3 for V in above equation.

m=(1.0×103kg/m3)(4.2×109m3)m=4.2×106kg

From equation (3) the final equation is,

mg=(3.10×104rv+0.870r2v2)N

Substitute 1.00mm for r , 4.2×106kg for m and 9.8m/s2 for g in the above equation.

(4.2×106kg)(9.8m/s2)=(3.10×104(1.00mm(103m1mm))v+0.870(1.00mm(103m1mm))2v2)N 41.16×106=(3.10×104(1.00×103m)v+0.870(1.00×103m)2v2)41.16×106=(3.10×107v+0.870×106v2)

The contribution of the first term of v to the air resistance is very small and it is ignored. The new equation is,

41.16×106=(0.870×106v2)v2=41.16×1060.870×106m/secv=41.16×1060.870×106m/secv6.87m/sec

Conclusion:

Therefore, the terminal speed for water droplets falling under their own weight in air for the drop radii equal to 1.00mm is 6.87m/sec .

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Chapter 6 Solutions

Physics for Scientists and Engineers, Volume 1

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