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nikklg [1K]
3 years ago
9

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressu

re of 4.5 × 105 Pa and a speed of 2.3 m/s. However, on the second floor, which is 4.0 m higher, the speed of the water is 4.4 m/s. The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?
Physics
1 answer:
sweet [91]3 years ago
8 0

Answer:

the gauge pressure of the water on the second floor is Pg2 = 403765 Pa = 4.03 * 10⁵ Pa

Explanation:

According to Bernoulli's equation ( conservation of mechanical energy, therefore we neglect the friction inside the pipes)

P1 + ρ*g*h1 + 1/2*ρ*v1² = P2 + ρ*g*h2 + 1/2*ρ*v2²

where ρ= density of water , h=height , P= absolute pressure, g= gravity ( 9.8 m/s²)

since the gauge pressure Pg is

Pg = P - Pa (atmospheric pressure)

Pg1 +Pa + ρ*g*h1 + 1/2*ρ*v1² = Pg2 + Pa + ρ*g*h2 + 1/2*ρ*v2²

Pg1 + ρ*g*h1 + 1/2*ρ*v1² = Pg2 + ρ*g*h2 + 1/2*ρ*v2²

since our reference point to measure heights can be chosen, we choose h1=0 and h2= 4m. Thus

Pg1 + 1/2*ρ*v1² = Pg2 + ρ*g*h2 + 1/2*ρ*v2²

therefore

Pg2 = Pg1 + 1/2*ρ*v1² -ρ*g*h2 - 1/2*ρ*v2² =  Pg1 -ρ*g*h2 -1/2*ρ*(v2²-v1²)

replacing values and assuming ρ=1000 kg/m³

Pg2 = Pg1 -ρ*g*h2 -1/2*ρ*(v2²-v1²) = 4.5 * 10⁵ Pa - 1000kg/m³*9.8 m/s²*4m -1/2* 1000 kg/m³ * [(4.4 m/s)² - (2.3 m/s)² ] = 403765 Pa = 4.03 * 10⁵ Pa

Pg2 =4.03 * 10⁵ Pa

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The work done by the normal force n when the box slides down a frictionless incline and gaining speed is zero.

<h3>What is normal force?</h3>

The force of contact is called the normal force. When the two surfaces are in contact with each other, then the normal force acts.

This force is applied by the solid bodies on each other in order to prevent the passing through each other.

A box slides down a frictionless incline, gaining speed. For this box, the value of work done by normal force has to be found out. Let's analyze the given condition.

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