As you take the stoppered part of the tube up the staircase you begin to see the water level drop around the 4th floor. As you continue up it does not continue up with you but stays at a constant level. What does that mean?
a. The pressure in the tubing is equal to the barometric pressure.
b. The tubing was unable to supply any more water to the tube for use.
c. The pressure outside the tube is higher that the water pressure inside the tube.

Answers

Answer 1

Answer:

a. The pressure in the tubing is equal to the barometric pressure.

Explanation:

Since in the question it is mentioned that the if you take the stoppert part of the tube than the level of warer would be fall approx 4th floor and if it is continued than it wont be continue but remains constant.

Now here first we do that the tube i.e. connected to the bucket should be taken up. In the first instance, the bucket supplies the water to the tube but it would not increased far away to the level of the barometric pressure

Hence, the correct option is a.


Related Questions

The engine of a model airplane must both spin a propeller and push air backward to propel the airplane forward. Model the propeller as three 0.30-m-long thin rods of mass 0.040 kg each, with the rotation axis at one end.
What is the moment of inertia of the propeller?
How much energy is required to rotate the propeller at 5800 rpm? Ignore the energy required to push the air.

Answers

Solution :

Given :

Length of the propeller rods, L =0.30 m

Mass of each, M = 0.040 kg

Moment of inertia of one propeller rod is given by  

[tex]$I=\frac{1}{3}\times M \times L^2$[/tex]

Therefore, total moment of inertia is

[tex]$I=3 \times \frac{1}{3}\times M \times L^2$[/tex]

[tex]$I=M\times L^2$[/tex]

[tex]$I=0.04\times (0.3)^2$[/tex]

  [tex]$0.0036 \ kg \ m^2$[/tex]

Now energy required is given by

[tex]$E=\frac{1}{2}\times I \times \omega^2 $[/tex]

where, angular speed, ω = 5800 rpm

[tex]$\omega = 5800 \times \frac{2 \pi}{60} $[/tex]

 = 607.4 rad/s

Therefore energy,

[tex]$E=\frac{1}{2}\times 0.0036 \times (607.4)^2 $[/tex]

   = 664.1 J

The moment of inertia of the propeller is 0.0036 kgm² and the energy required is 663.21 J

Energy required for propeller:

Given that the mass of the propellers is m = 0.040kg,

and their length is L = 0.30m

The moment of inertia of a rod with the rotation axis at one end is given by :

[tex]I = \frac{1}{3}m L^2[/tex]

so for 3 propellers:

[tex]I=3\times\frac{1}{3}\times(0.04)\times(0.3)^2[/tex]

I = 0.04 × 0.09

I = 0.0036 kgm²

Now, the frequency is given f = 5800 rpm

so anguar speed, ω = 5800×(2π/60)

ω = 607 rad/s

Energy required:

E = ¹/₂Iω²

E = 0.5 × 0.0036 × (607)² J

E = 663.21 J

Learn more about moment of inertia:

https://brainly.com/question/15248039?referrer=searchResults

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