In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 pa. the water leaves the bottom of the tank through a nozzle that is directed straight upward. the opening of the nozzle is 2.86 m below the surface of the water. (a) find the speed at which the water leaves the nozzle. (b) ignoring air resistance and viscous effects, determine the height (relative to the submerged end of nozzle) to which the water rises.

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot,  at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot  = Pgas

        = 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵  =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵  = 1000 x 9.81 x h

h = 65.85 meters

answer///terminator or twilight zone;

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle