Question

fall 25 home announcements syllabus modules grades honorlock people course reserves library resources microsoft education question 1 1 pts on a balanced seesaw, a boy three times as heavy as his partner sits more than 1/3 the distance from the fulcrum. less than 1/3 the distance from the fulcrum. 1/3 the distance from the fulcrum. question 2 1 pts if a turntables rotational speed is doubled, then the linear speed of a pet hamster sitting on the edge of the record will remain the same.

Explanation:

Question 1

Step1: Recall torque - balance condition

For a balanced seesaw, the torques on both sides of the fulcrum are equal. Torque $\tau = F\times d$, where $F$ is the force (proportional to mass) and $d$ is the distance from the fulcrum. Let the mass of the partner be $m$ and the distance of the partner from the fulcrum be $d_1$, and the mass of the boy be $M = 3m$ and the distance of the boy from the fulcrum be $d_2$. Since $\tau_1=\tau_2$, we have $m\times d_1=3m\times d_2$. Solving for $d_2$, we get $d_2=\frac{1}{3}d_1$.

Step1: Use the relationship between linear and rotational speed

The relationship between linear speed $v$ and rotational speed $\omega$ is given by $v = r\omega$, where $r$ is the radius (distance from the axis of rotation). If the rotational speed $\omega$ is doubled ($\omega_{new}=2\omega_{old}$) and $r$ remains the same (the hamster is still at the edge of the turntable, so $r$ is constant), then using the formula $v = r\omega$, the new linear speed $v_{new}=r\omega_{new}=r\times2\omega_{old}=2v_{old}$.

Answer:

The boy should sit $\frac{1}{3}$ the distance from the fulcrum.

Question 2