Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 9, Problem 102P

(a)

To determine

The weight of beach ball including the air inside it.

(a)

Expert Solution
Check Mark

Answer to Problem 102P

The weight of beach ball is 1.4N.

Explanation of Solution

Radius of ball is 20.0cm, mass of ball is 0.10kg, and the air density is 1.3kg/m3.

Write the equation for the weight of ball with air inside it.

W=mbg+43πr3ρag

Here, W is the weight of ball with air, mb  is the mass of ball, g is the acceleration due to gravity, r is the radius of ball, and ρa is the density of air.

Conclusion:

Substitute 0.10kg for mb, 9.8m/s2 for g, 20.0cm for r, 3.14 for π, and 1.3kg/m3 for ρa in the above equation to find W.

W=((0.10kg)(9.8m/s2))+(43(3.14)(20.0cm(1m100cm))3(1.3kg/m3)((9.8m/s2)))=0.98N+0.12N=1.4N

Therefore, the weight is 1.4N.

(b)

To determine

The buoyant force on beach ball.

(b)

Expert Solution
Check Mark

Answer to Problem 102P

The buoyant force is 0.43Ninupwarddirection.

Explanation of Solution

Radius of ball is 20.0cm, mass of ball is 0.10kg, and the air density is 1.3kg/m3.

Write the equation for buoyant force on each ball.

FB=ρag(43πr3)

Here, FB is the buoyant force.

Conclusion:

Substitute 9.8m/s2 for g, 20.0cm for r, 3.14 for π, and 1.3kg/m3 for ρa in the above equation to find FB.

FB=(1.3kg/m3)(9.8m/s2)43(3.14)(20.0cm(1m100cm))3=0.43N

Buoyant force will acts in opposite direction of weight. That is directed upwards.

Therefore, the buoyant force is 0.43Ninupwarddirection.

(c)

To determine

The acceleration of ball at the top of its path.

(c)

Expert Solution
Check Mark

Answer to Problem 102P

The acceleration is 6.8m/s2indownwarddirection.

Explanation of Solution

Radius of ball is 20.0cm, mass of ball is 0.10kg, and the air density is 1.3kg/m3.

The free body diagram is shown below in figure 1.

Physics, Chapter 9, Problem 102P

Write the equation net force on the ball.

ma=FBmg

Here, a is the acceleration of ball at the top of its path and m is the mass of ball with air.

Rewrite the above equation in terms of a.

a=FBmgm=FBmg

Write the equation to find m.

m=mb+ρa(43πr3)

Rewrite the equation for a by substituting the above relation for m and ρag(43πr3) for FB.

a=FBmgm=ρag(43πr3)mb+ρa(43πr3)g=g(mbρa(43πr3)+1)g=g(1+(mbρa(43πr3))11)

Conclusion:

Substitute 9.8m/s2 for g, 0.10kg for mb, 20.0cm for r, 3.14 for π, and 1.3kg/m3 for ρa in the above equation to find FB.

a=(9.8m/s2)(1+(0.10kg(1.3kg/m3)43(3.14)(20.0cm(1m100cm))3)11)=(9.8m/s2)(0.694)=6.8m/s2

Negative value indicates that the acceleration is directed in downward direction.

Therefore, the acceleration is 6.8m/s2indownwarddirection.

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