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

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it