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In 1994, Susan Williams of California blew a bubble-gum bubble with a diameter of 58.4 cm. If this bubble were rigid and the centripetal acceleration of the equilateral points of the bubble were 8.5*10^-2 m/s^2, what would the tangential speed of those points be?

Answer :

skyluke89

Answer:

0.158 m/s

Explanation:

The centripetal acceleration of the equilateral points of the bubble is given by

[tex]a=\frac{v^2}{r}[/tex]

where

v is the tangential speed

r is the radius of the bubble

In this problem, we have:

[tex]a=8.5\cdot 10^{-2} m/s^2[/tex] is the centripetal acceleration

[tex]d=58.4 cm[/tex] is the diameter of the bubble, so the radius is

[tex]r=\frac{58.4 cm}{2}=29.2 cm=0.292 m[/tex]

Therefore, we can re-arrange the previous equation to find the tangential speed of the equilateral points:

[tex]v=\sqrt{ar}=\sqrt{(8.5\cdot 10^{-2} m/s^2)(0.292 m)}=0.158 m/s[/tex]

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