Use bernoulli's equation to find out how fast water leaves an opening in a water tank. the water level is 0.75 m above the opening. the top of the water tank is opened to the atmosphere.

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

answer would be d.) i hope this

explanation: