Answer:
3.84 m/s
Explanation:
Using Bernoulli's equation below:
P1 + (1/2ρv1²) + h1ρg = P2 + (1/2ρv2²) + h2ρg
where P1 = P2 atmospheric pressure
(1/2ρv1²) + h1ρg = (1/2ρv2²) + h2pg
collect the like terms
h1ρg - h2ρg = (1/2ρv2²) - (1/2ρv1²)
factorize the expression by removing the like terms on both sides
gρ(h1 - h2) = 1/2ρ( v2² - v1²)
divide both side by rho (density in kg/m³, ρ )
g(h1 - h2) = 1/2 (v2² - v1²)
assuming the surface of the tank is large and the speed of water then at the tank surface v1 = 0
2g(h1 - h2) = v2²
take the square root of both side and h1 - h2 is the difference between the surface of the tank and the opening where water is coming out in meters
√2g(h1 - h2) = √ v2²
v2 = √2g(h1-h2) = √ 2 × 9.81×0.75 = 3.84 m/s
