Answer:
28.97 m/s
Explanation:
height of water (H1) = 10.7 m
Gauge pressure (Pg) = 3.2 x 1.01325 x 10^{5} = 324240 N/ m^[2}
pressure at the top (P1) = gauge pressure (Pg) + atmospheric pressure(Pa) (Pg + Pa)
acceleration due to gravity (g) = 9.8 m/s^{2}
height at the bottom (H2) = 0
pressure at the bottom (P2) = atmospheric pressure = (Pa)
density of sea water (ρ) = 1030 kg/m^{3}
speed of water at the top (V1) = 0 (assuming the tank is big, the speed of water at any point on the surface is almost 0 and also the hole is very small compares to the size of the tank)
speed of water at the top (V2) = ?
applying Bernoulli's equation we can find the value of V2.
[tex]P1+pgh1+\frac{1}{2}pV1^{2}=P2+pgh2+\frac{1}{2}.pV2^{2}[/tex] (take note that [tex]p[/tex] represents ρ whcih is the density of sea water)
recall that P1 = Pg + Pa and P2 = Pa
[tex]Pg + Pa + pgh1 + \frac{1}{2}pV1^{2} = Pa + pgh2 + \frac{1}{2}.pV2^{2}[/tex]
[tex]Pg + pgh1 + \frac{1}{2}pV1^{2} = pgh2 + \frac{1}{2}pV2^{2}[/tex]
[tex]324240 + (1030 x 9.8 x 10.7) + (\frac{1}{2}x1030x0^{2}) = (1030x9.8x0) + (\frac{1}{2}x1030xV2^{2}[/tex])
[tex]324240 + (1030 x 9.8 x 10.7) = (\frac{1}{2}x1030xV2^{2}[/tex])
V2 =
[tex]324240 + (1030 x 9.8 x 10.7) = \frac{1}{2}x1030xV2^{2}[/tex]
V2 = [tex]\sqrt{\frac{324240 + (1030 x 9.8 x 10.7)}{ \frac{1}{2}x1030}}[/tex]
V2 = 28.97 m/s
The water was moving out from the bottom at a velocity of 29 m/s
Applying Bernoulli's equation:
P + ρgh + (1/2)ρv² = constant
P is pressure, g is acceleration due to gravity, h is height, v is velocity, ρ is density.
At top:
P = 3.20 atm = 3.20 * 101300 Pa, ρ = 1025 kg/m³, v = 0, g = 9.81 m/s, h = 10.7 m, hence:
P + ρgh + (1/2)ρv² = ( 3.20 * 101300) + (1025 * 9.81 * 10.7) + 0
At bottom:
h = 0, P = 0
P + ρgh + (1/2)ρv² = 0 + 0 + (1/2* 1025)v²
Equating top and bottom:
( 3.20 * 101300) + (1025 * 9.81 * 10.7) = (1/2* 1025)v²
v = 29 m/s
The water was moving out from the bottom at a velocity of 29 m/s
Find out more at: https://brainly.com/question/23841792
