**Question 6:**Two kids are sitting on a seesaw and the seesaw is balanced. Kid 1 has a mass of 35 kg and kid 2 has a mass of 42 kg. Kid 1 is sitting 0.5 m away from the center of the seesaw. How far away is kid 2 from the center of the seesaw?**Explanation:**To solve this problem, we utilize the principle of moments (torque) which states that for the seesaw to be balanced, the moments (force times distance) on both sides must be equal.We have:- Mass of kid 1 (m1) = 35 kg- Distance of kid 1 from the center of the seesaw (d1) = 0.5 m- Mass of kid 2 (m2) = 42 kgLet \( d_2 \) be the distance of kid 2 from the center of the seesaw. The moment caused by kid 1 is \( m_1 \times d_1 \) and the moment caused by kid 2 is \( m_2 \times d_2 \). For the seesaw to be balanced:\[ m_1 \times d_1 = m_2 \times d_2 \]Substituting the given values:\[ 35 \, \text{kg} \times 0.5 \, \text{m} = 42 \, \text{kg} \times d_2 \]Solving for \( d_2 \):\[ 17.5 = 42 \, d_2 \]\[ d_2 = \frac{17.5}{42} \]\[ d_2 \approx 0.4167 \, \text{m} \]Therefore, kid 2 is approximately 0.4167 meters away from the center of the seesaw.

Given:- m1 = 35 Kg m2 = 42 Kg x1 = 0.5 m (from the center) To find:- x2

6. Two kids are sitting on a seesaw and the seesaw is balanced. Kid 1 has a mass of 35 kg and kid two has a mass of 42 kg. Kid 1 is sitting 0.5 m away from the center of the seesaw. How far away is kid 2 from the center of the seesaw?

6. Two kids are sitting on a seesaw and the seesaw is balanced. Kid 1 has a mass of 35 kg and kid two has a mass of 42 kg. Kid 1 is sitting 0.5 m away from the center of the seesaw. How far away is kid 2 from the center of the seesaw?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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