the edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. the centripetal…
the edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. the centripetal acceleration of the edge of the disc is m/s².\n0.0051\n0.43\n25\n84
Answer
Explanation:
Step1: Recall centripetal - acceleration formula
The formula for centripetal acceleration is $a_c=\frac{v^2}{r}$, where $v$ is the tangential speed and $r$ is the radius.
Step2: Substitute given values
Given $v = 3.3\ m/s$ and $r=0.13\ m$. Substitute into the formula: $a_c=\frac{(3.3)^2}{0.13}$. First, calculate $(3.3)^2 = 10.89$. Then, $a_c=\frac{10.89}{0.13}\approx84\ m/s^2$.
Answer:
84