Answer:
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
