Question

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?

Answer 1

0.605 m (approx.)

Step-by-step explanation:

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

`a = (v^2)/(r)`

From this equation, we can solve for
`$r$` like this,

`r = (v^2)/(a)\\r = (0.51^2)/(0.43) m \\ = 0.605 m`

Answer 2

0.84m

Step-by-step 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