Question

"A hole is punched in a full milk carton, 10 cm below the top. What is the initial velocity of outflow?"

Answer 1

P 1 = 101,325 Pa (atmospheric pressure)
Milk has almost same density as water: (Rho)= 1,000 kg /m³ 
P 2 = 101,325 Pa + 1,000 kg/m³ · 9.81 m/s² · 0.1 m = 102,306 Pa
The hydrostatic equation:
P 1 + (Rho)v1² / 2 = P 2 + (Rho)·g·h2
101,325 + 1,000 v1²/2  = 102,306 + 1000 · 9.81 · 0.1
500 v 1² = 102,306 + 981 - 101,325
v 1² = 3.924
v 1 = √ 3.924
v 1 = 1.98 m/s
The initial velocity of outflow is 1.98 m/s.

Answer 2

Answer:

The initial velocity of outflow is 1.4 m/s.

Explanation:

As a hole is punched in a full milk carton, and we have to calculate the initial velocity of outflow 10 cm below the top. We use the concept of conservation of energy.  

Further Explanation:

Using the conservation of energy, below the top the potential energy converted in kinetic energy

P.E =K.E

$mgh=\frac{1}{2} mv^{2}$

As mass of milk does not change at the top and 10 cm below the top, therefore  

$v=\sqrt{2gh}$

Here, v is the initial velocity,  

g is acceleration due to gravity  

h is height.  

Given: h = 10 cm and g = 9.8 m/s2

Substituting the given values, we get

$v=\sqrt{2×9.8×0.1m}$  

$v=\sqrt{1.96}=1.4 m/s$

Lear more:

brainly.com/question/1581227

Key word:  

Conservation of energy, Kinetic energy, potential energy.