Respuesta :
Solution:
Given:
height of geyser, h = 50 m
speed of water at ground can be given by third eqn of motion with v = 0 m/s:
[tex]u^{2} = v^{2} + 2gh[/tex]
putting v = 0 m/s in the above eqn, we get:
u = [tex]\sqrt{2gh}[/tex] (1)
Now, pressure is given by:
[tex]\Delta p = \frac{1}{2}u^{2}\rho \\[/tex] (2)
where,
[tex]\Delta p[/tex] = density of water = 1000 kg/[tex]m^{3}[/tex]
g = 9.8 m/[tex]s^{2}[/tex]
Using eqn (1) and (2):
[tex]\Delta p = \frac{1}{2}(\sqrt{2gh})^{2}\rho[/tex]
⇒ [tex]\Delta p = \frac{1}{2}\rho gh[/tex]
[tex]\Delta p = \frac{1}{2}\times2\times 1000\times 9.8\times 50[/tex] = [tex]\Delta p[/tex] = 490000 Pa
[tex]p_{atm}[/tex] = 101325 Pa
For absolute pressure, p:
p = [tex]\Delta p[/tex] + [tex]p_{atm}[/tex]
p = 490000 - 101325
p = 388675 Pa
Therefore, pressure in hot springs for the water to attain the height of 50m is 388675 Pa
