In a very large closed tank, the absolute pressure of the air above the water is 5.27 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 5.39 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.

30.95 m/s

48.83 m

Explanation:

The Pressure difference that is present across nozzle is given as

∆P = (inside) - (outside)

ΔP = [5.27*10^5 + hρg] - 1 atm

∆P = 5.27*10^5 + [5.39 x 1000 * 9.8] - [1.01*10^5]

ΔP = 5.27*10^5 + 5.282*10^4 - 1.01*10^5

ΔP = 5.79*10^5 - 1.01*10^5

∆P = 4.79*10^5 Pa

Making use of the Bernoulli horizontal flow equation

∆P = ½ρ(v2² - v1²), where v1 = 0 (water in tank), this makes the equation

∆P = ½ρ(v2)², solving for v2, we have

v2 = √(2.∆P/ρ)

v2 = √[(2 * 4.79*10^5)/1000]

v2 = √958

v2 = 30.95 m/s

B

Using law of conservation of energy, we have

½mv² = mgh

½v² = gh, making h subject of formula

h = v² /2g

h = (30.95)² / (2 * 9.81)

h = 958 / 19.62

h = 48.83 m

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )