Suppose a merry-go-round is spinning once every 3 seconds. If a point on the outside edge has a speed of 12.56 feet per second, what is the radius of the merry-go-round? (Use 3.14 for

My account
Edit my Profile
Private messages
My favorites

Ask a Question
Questions
Tags
Categories
Ask a Question

Ibolit asked Jun 9, 2023

474,053 views

36 votes

Suppose a merry-go-round is spinning once every 3 seconds. If a point on the outside edge has a speed of 12.56 feet per second, what is the radius of the merry-go-round? (Use 3.14 for

Mathematics
high-school

Ibolit asked Jun 9, 2023

by Ibolit

3.2k points

0 Comments

Your comment on this question:

Email me at this address if a comment is added after mine:Email me if a comment is added after mine

Privacy: Your email address will only be used for sending these notifications.

Your answer

Email me at this address if my answer is selected or commented on:Email me if my answer is selected or commented on

Privacy: Your email address will only be used for sending these notifications.

1 Answer

14 votes

Answer:

r = 6 feet

Explanation:

Ler r represents the radius

Hence perimeter = edge speed times 3second

2 times 3.14 times r = 12.56 times 3

r = 6 feet

Raghavendra Bankapur answered Jun 13, 2023

by Raghavendra Bankapur

2.6k points

0 Comments

Your comment on this answer:

Email me at this address if a comment is added after mine:Email me if a comment is added after mine

Privacy: Your email address will only be used for sending these notifications.

Ask a Question

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

1.6m questions

2.0m answers

0 Comments

Your comment on this question:

Your answer

1 Answer

0 Comments

Your comment on this answer:

Other Questions