a spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. if the top has a radius of…

a spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. if the top has a radius of 0.10 m, what is the centripetal acceleration of the edge of the top?\n0.016 m/s²\n0.73 m/s²\n4.0 m/s²\n6.3 m/s²

a spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. if the top has a radius of 0.10 m, what is the centripetal acceleration of the edge of the top?\n0.016 m/s²\n0.73 m/s²\n4.0 m/s²\n6.3 m/s²

Answer

Explanation:

Step1: Recall centripetal - acceleration formula

The formula for centripetal acceleration is $a_c=\frac{v^{2}}{r}$, where $v$ is the linear - speed and $r$ is the radius.

Step2: Substitute given values

Given $v = 0.63\ m/s$ and $r=0.10\ m$. Substitute these values into the formula: $a_c=\frac{(0.63)^{2}}{0.10}$. First, calculate $(0.63)^{2}=0.63\times0.63 = 0.3969$. Then, $a_c=\frac{0.3969}{0.10}=3.969\approx4.0\ m/s^{2}$.

Answer:

C. $4.0\ m/s^{2}$