A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both
2 answers:
D. Nothing will happen; the seesaw will still be balanced.
<h3>Further explanation </h3>The force acting on a system with static equilibrium is 0
(forces acting as translational motion only, not including rotational forces)
For objects undergoing rotation, the equilibrium must be met
A heavy boy (Hb) and a lightweight girl (Lg) are balanced on a mass-less seesaw
Because there is a balance of rotation, the torque equation:
Στ = 0
Hb.r1-Lg.r2 = 0
Hb.r1 = Lg.r2 (equation 1)
If they both move forward so that they are one-half their original distance from the pivot point, then the distance of the two children to the pivot point is reduced to half
Then the torque equation:
This equation remains the same as equation 1, so the seesaw will still be balanced.
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Answer:
distance difference would a) increase
speed difference would f) stay the same
Explanation:
Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.
Their equations of motion for distance and velocities are
Their difference in distance are therefore:
(As
So as time progress t increases, Δs would also increases, their distance becomes wider with time.
Similarly for their velocity difference
Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.
This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.