The water molecules with a slower speed are escaping
Answer:
w = 5832.372 Joules
Explanation:
Mass of water, m = 20 kg
The water was pulled up to a height of 35 meters, i.e. h = 35 m
It takes 14 minutes to pull up the water through the height, 35 m
speed = distance/ time = 35/14 = 2.5 m/min
The bucket's height, y = speed * time = 2.5t meters
6 kg of water drips out of the bucket throughout the 14 minutes
The rate at which the water drips drips out = (6/14) = 0.4286 kg/min
Mass of water that drips out in time, t = 0.4286t kg
The mass of water remaining = (20 - 0.4286t) kg
Change in Workdone, Δw = mgΔy
Δy = 2.5 Δt
Δw = mg * 2.5 Δt
dw = (20 - 0.4286t)g2.5 dt
integrating both sides
dw = (50g - 1.07gt)dt
where b = 0, a = 14
w = 50gt - 1.07g(t²)/2 g = 9.8 m/s²
w = 490t - 5.243t²
w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)
w = 6860 - 1027.628
w = 5832.372 Joules
Answer:
Your answer is D It does not need to be repeatable.
the reason for this is because C is correct.
You need to be able to experiment on something multiple times so that you can gather further data to imbedded your evidence in facts.
Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as

for uniform cylinder we will have

so we have


now we have




now we have


Part b)
speed of the bucket can be found using kinematics
so we have



Part c)
now in order to find the time of fall we can use another equation



Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance


