Gravity attracts an object to
1 answer:
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Answer:
-1.43 m/s relative to the shore
Explanation:
Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:
where are the mass of the swimmer and raft, respectively.
are the velocities of the swimmer and the raft after the run, respectively. We can solve for
So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore
Answer:
v = 7.67 m/s for L= 1m
Explanation:
Let's use the conservation of mechanical energy, at the highest point and the lowest point
Initial. Vertical ruler
Em₀ = mg h
Final. Just before touching the floor
= K = ½ I w²
Em₀ =
m g h = ½ I w²
The moment of inertia of a ruler that turns on one end is
I = 1/3 m L²
Let's replace
m g h = ½ (1/3 m L²) w²2
g h = 1/6 L² w²
They ask for the speed of the end so the height h is equal to the length of the ruler
g L = 1/6 L² w²
The linear and angular variables are related
v = w r
w = v / r
In this case the point of interest a in strangers r = L
g L = 1/6 L² v² / L²
v = √ 6 g L
Let's calculate
Assume that the length of the meter is L = 1 m
v = √ (6 9.8 1)
v = 7.67 m/s