Estimate the number of Ping-Pong balls that can be packed into an average size room (without crushing them). Given that Ping-Pong ball has a radius of about 1.5 cm and assume that a typical room has dimensions12 ft ×18 ft ×9 ft.

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cryssatemp cryssatemp · Answer 1 · 2019-09-05T06:27:29+02:00

Answer: 3,893,845.918 Ping-Pong balls

Explanation:

The volume of an average room is:

[tex]V_{room}=(length)(width)(height)[/tex] (1)

[tex]V_{room}=(12 ft)(18 ft)(9 ft)=1944 ft^{3}[/tex] (2)

Now let’s transform this [tex]V_{room}[/tex] to units of [tex]cm^{3}[/tex], knowing [tex]1 ft=30.48 cm[/tex]:

[tex]V_{room}=1944 ft^{3}\frac{{(30.48 cm)}^{3}}{1ft^{3}}=55,047,949.78 cm^{3}[/tex] (3)

On the other hand, we have Ping-Pong balls with a radius [tex]r=1.5 cm[/tex], and their volume is given by:

[tex]V_{balls}=\frac{4}{3} \pi r^{3}[/tex] (4)

[tex]V_{balls}=\frac{4}{3} \pi (1.5 cm)^{3}[/tex] (5)

[tex]V_{balls}=14.137 cm^{3}[/tex] (6)

Now, the number [tex]n[/tex] of Ping-Pong balls that can be packed into the room is:

[tex]n=\frac{V_{room}}{V_{balls}}[/tex] (7)

[tex]n=\frac{55,047,949.78 cm^{3}}{14.137 cm^{3}}[/tex] (8)

[tex]n=3,893,845.918[/tex] This is the number of Ping-Pong balls that can be packed into an average size room