We appreciate your visit to A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle At some point its speed is. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Final answer:
The stone is at the bottom of the vertical circle due to the tension being greater than the weight of the stone plus the required centripetal force to maintain speed.
Explanation:
This is a Physics problem related to circular motion. Given that the weight of the stone (mg) is 2 kg * 9.8 m/s^2 = 19.6 N and the tension in the string is 118 N, we know that the net force on the stone is directed upwards. When we consider that the centripetal force needed to maintain speed in the circular path is mv^2 / r (mass * velocity^2 / radius), in this case, 2 kg * (7 m/s)^2 / 1 m = 98 N. By observed that the tension is higher than the weight of the stone plus the required centripetal force (19.6N + 98N), it is clear that the stone is at the bottom of the circle.
Learn more about Circular Motion here:
https://brainly.com/question/35892634
#SPJ11
Thanks for taking the time to read A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle At some point its speed is. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
