Respuesta :

This is a problem of volume. So we have the following volumes:

1. For the balls: Volume of a sphere.

[tex]V_{1s} = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (2.7)^{3} = 82.44in^{3}[/tex]

Given that we have three balls, then the total volume of the balls is:

[tex]V_{3s} = 3\times 82.44in^{3}=247.34in^{3}[/tex]

1. For the cylindrical can:

[tex]V_{c}=\pi r^{2}h[/tex]

h is the height, that is, if the balls touch the sides, bottom and top of the canister, then it is true that:

[tex]h=6r=6(2.7)=16.2in[/tex]

The radius of the cylindrical can is the same as the radius of a ball, therefore:

[tex]V_{c}=\pi (2.7)^{2}(16.2)=371.01in^{3}[/tex]

Finally the volume of the air between the tennis balls in the can is:

[tex]V_{A}=V_{c}-V_{3s}=371.01-247.34 \rightarrow \boxed{V_{A}=123.67in^{3}}[/tex]


Answer: 123.67 in³

Explanation:

1) State the problem in your own word.

Calcualte the volume of the air in a tube with 3 tennis ball whose radius is 2.7 cm, knowing that the walls of the tube touch the balls, so as the base and the cap.

2) Find the volumen of the tube without balls:

i) Formula for the volume of a cylinder V = π × r² × h, where r is the radius of the cylinder and h is the height.

Here r = radius of a ball = 2.7 in, and h = diameter of 3 balls = 3×2.7in × 2 = 16.2 in

ii) Volume of the tube = π (2.7in)² (16.2in) = 371.02 in³

iii) Formula for the volume of a bal (a sphere) : V = (4/3) π r³

iv) Volume of one ball = (4/3)π (2.7in)³ = 82.45 in³

v) Volume of 3 balls = 3×82.42in³ = 247.35 in³

vi) Volume of air bettween the tennis ball and the can = volume of the tube - volume of the 3 balls = 371.02 in³ - 247.35 in³ = 123.67 in³

That is the answer: 123.67 in³

And happy commencement!