Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is originally empty, what does the scale read (in newtons) 3.20 s after water starts to accumulate in it

F_scale ≈ 15.12 N

Explanation:

We are given;

Mass flow rate; m' = 0.2 l/s

Time; t = 3.2 s

Mass of bucket; m_b = 0.730 kg

Height; h = 3.6 m

Now, mass of water is;

m_w = 0.2 l/s × 3.2 s

m_w = 0.64 l

From conversion, 1 litre = 1 kg

Thus: m_w = 0.64 kg

Now, let's calculate final velocity from Newton's third equation of motion.

v² = u² + 2gh

Initial velocity is 0. Thus;

v² = 0 + 2(9.8 × 3.6)

v² = 70.56

v = √70.56

v = 8.4 m/s

Now, total mass of water and bucket is;

m_t = m_w + m_b = 0.64 + 0.73

m_t = 1.37 kg

Force on the scale is calculated from;

F_scale = (m_t)g + (m_w)v/t

F_scale = (1.37 × 9.81) + (0.64 × 8.4/3.2)

F_scale ≈ 15.12 N

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~