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
These forces form a force pair. Use Newton's third law, and you see that the trailer pulls back at with the same force. The answer is d.
Answer:
105.8 m
46 m/s
Explanation:
From the time the rocket is launched to the time it reaches its maximum height:
v = 0 m/s
a = -10 m/s²
t = 9.2 s / 2 = 4.6 s
Find: Δy and v₀
Δy = vt − ½ at²
Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²
Δy = 105.8 m
v = at + v₀
0 m/s = (-10 m/s²) (4.6 s) + v₀
v₀ = 46 m/s
Answer:
A. velocity has a direction .. .
with magnitude too but speed has only magnitude
Dispersion im pretty sure
