Answer:
v_f = 0.87 m/s
Explanation:
We are given;
F_avg = -17700 N (negative because it's backward)
m = 117 kg
Δt = 5.50 × 10^(−2) s
v_i = 7.45 m/s
Now, formula for impulse is given by;
I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s
From impulse momentum theory, we know that;
Change in momentum of particle is equal to impulse.
Thus,
Δp = I = m•v_f - m•v_i
Thus,
-973.5= 117(v_f - 7.45)
Thus,
-973.5/117 = (v_f - 7.45)
-8.3205 + 7.45 = v_f
v_f = - 0.87 m/s
We'll take absolute value as;
v_f = 0.87 m/s
Define
v = volume of a drop per second, cm³/s
The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.
Therefore,

The average volume of a single drop is approximately 0.0556 cm³.
Answer: 0.0556 cm³
I believe it would be 2m/s.