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A squirrel runs at a steady rate of 0.51 m/s in a circular path around a tree. If the squirrel's centripetal acceleration is 0.43 m/s 2 , what is the radius of the circle?

Respuesta :

corneltuttle

Answer:

0.84m

Explanation:

0.51+0.43=0.93

πr can be 22/7 or 3.14

radius is 2x2= 4

3.14 divide0.93divide 4

=0.84m/s

Answer:

0.605 m (approx.)

Explanation:

We know, given the radius and the circular velocity we can yield the centripetal acceleration using this equation,

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

From this equation, we can solve for [tex]$r$[/tex] like this,

[tex]r = \frac{v^2}{a}\\r = \frac{0.51^2}{0.43} m \\ = 0.605 m[/tex]