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.
Answer:
20.2 seconds
Explanation:
The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.
The crate is in free fall, so a = -9.8 m/s².
The crate falls downward, so Δx = -2000 m.
Find: t, the time it takes for the crate to land.
Δx = v₀ t + ½ at²
-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 20.2 s
It takes 20.2 seconds for the crate to land.
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