About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air.
Use g = 9.80 m/s2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m3.

(a) What is the speed of the water when it emerges from the ground? m/s

(b) Assuming the water travels to the surface through a narrow crack that extends 9.00 m below the surface, and that the water comes from a chamber with a large cross-sectional area, what is the pressure in the chamber?

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cryssatemp cryssatemp · Answer 1 · 2019-12-01T14:43:06+01:00

Answers:

a) [tex]8820 m/s[/tex]

b) [tex]189500 Pa[/tex]

Explanation:

We have the following data:

[tex]t=30 min \frac{60 s}{1 min}=1800 s[/tex] is the time

[tex]h=11 m[/tex] is the height the water reaches vertically

[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity

[tex]P_{air}=101.3 kPa=101.3(10)^{3} Pa[/tex] is the pressure of air

[tex]\rho_{water}=1000 kg/m^{3}[/tex] is the density of water

Knowing this, let's begin:

Here we will use the following equation:

[tex]h=h_{o}+V_{o}t-\frac{g}{2}t^{2}[/tex] (1)

Where:

[tex]h_{o}=0 m[/tex] is the initial height of water

[tex]V_{o}[/tex] is the initial speed of water

Isolating [tex]V_{o}[/tex]:

[tex]V_{o}=\frac{1}{t}(h+\frac{g}{2}t^{2})[/tex] (2)

[tex]V_{o}=\frac{1}{1800 s}(11 m+\frac{9.8 m/s^{2}}{2}(1800 s)^{2})[/tex]

[tex]V_{o}=8820.006 m/s \approx 8820 m/s[/tex] (3)

In this part we will use the following equation:

[tex]P=\rho_{water} g d + P_{air}[/tex] (4)

Where:

[tex]P[/tex] is the absolute pressure in the chamber

[tex]d=9 m[/tex] is the depth

[tex]P=(1000 kg/m^{3})(9.8 m/s^{2})(9 m) + 101.3(10)^{3} Pa[/tex]

[tex]P=189500 Pa[/tex] (5)