Question

A waterwheel built in hamah, syria, has a radius of 20.0 m. if the tangential velocity at the wheel's edge is 7.85 m/s, what is the centripetal acceleration of the wheel?

Answer 1

The centripetal acceleration of the waterwheel is approximately 3.08 m/s^2.

Step-by-step explanation:

To find the centripetal acceleration of the waterwheel, we can use the formula:
centripetal acceleration = tangential velocity^2 / radius

Given that the radius of the waterwheel is 20.0 m and the tangential velocity is 7.85 m/s, we can substitute these values into the formula:
centripetal acceleration = (7.85 m/s)^2 / 20.0 m

Simplifying this equation gives us:
centripetal acceleration ≈ 3.08 m/s^2

Therefore, the centripetal acceleration of the waterwheel is approximately 3.08 m/s^2.

Answer 2

We are tasked to solved for the centripetal acceleration given the radius and the tangetial velocity. USing the formula of Centripetal acceleration,


`a_(c)` =
`( v_(t) ^(2) )/(r)`

Given:

`v_(t)` = 7.85 m/s
r= 20.0 m

By Substitution,


`a_(c)` =
`( v_(t) ^(2) )/(r)`

`a_(c)` =
`( 7.85^(2) )/(20)`

`a_(c)` =
`( 61.6225)/(20)`

`a_(c)` = 3.081125 m/
`s^(2)`

Therefore, the centripetal acceleration is 3.08 m/
`s^(2)`