Question

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?

Answer 1

Answers:

a)
`8820 m/s`

b)
`189500 Pa`

Step-by-step explanation:

We have the following data:

`t=30 min (60 s)/(1 min)=1800 s` is the time

`h=11 m` is the height the water reaches vertically

`g=9.8 m/s^(2)` is the acceleration due gravity

`P_(air)=101.3 kPa=101.3(10)^(3) Pa` is the pressure of air

`\rho_(water)=1000 kg/m^(3)` is the density of water

Knowing this, let's begin:

a) Initial speed of water

Here we will use the following equation:

`h=h_(o)+V_(o)t-(g)/(2)t^(2)` (1)

Where:

`h_(o)=0 m` is the initial height of water

`V_(o)` is the initial speed of water

Isolating
`V_(o)`:

`V_(o)=(1)/(t)(h+(g)/(2)t^(2))` (2)

`V_(o)=(1)/(1800 s)(11 m+(9.8 m/s^(2))/(2)(1800 s)^(2))`

`V_(o)=8820.006 m/s \approx 8820 m/s` (3)

b) Pressure in the chamber

In this part we will use the following equation:

`P=\rho_(water) g d + P_(air)` (4)

Where:

`P` is the absolute pressure in the chamber

`d=9 m` is the depth

`P=(1000 kg/m^(3))(9.8 m/s^(2))(9 m) + 101.3(10)^(3) Pa`

`P=189500 Pa` (5)