Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

A 50.0-kg child stands at the rim of a merry-go-round of radius 2.25 m, rotating with an angular speed of 3.30 rad/s.. What is the child's centripetal acceleration?

Sagot :

tutorAnne

Answer: the child's centripetal acceleration=24.50 m/s²

Explanation:

Given that mass of child= 50 kg

radius of merry go round= 2.25m

angular speed = 3.30 rad/s

 

Centripetal Acceleration  = v²/ r

  But  V= ωr

So Centripetal Acceleration  = v²/ r =  (ωr)²/ r

=(3.30)² x  (2.25)²/ 2.25 = (3.30)² x  2.25

=24.5025m/s²

=24.50 m/s²

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.