Answer:
a) The speed is 61.42 m/s
b) The drag force is 10.32 N
Explanation:
a) The Reynold´s number for the model and prototype is:


Equaling both Reynold's number:

Clearing Vm:

b) The drag force is:

Vapor pressure<span> or equilibrium </span>vapor pressure<span> is defined as the </span>pressure<span> exerted by a </span>vapor<span> in thermodynamic equilibrium with its condensed phases at a certain temperature. It is independent with atmospheric pressure since it does not change by changing the atmospheric pressure only. </span>
W = 25 J
Explanation:
Work done on an object is defined as

Answer:
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Answer:
P₁ = 2.3506 10⁵ Pa
Explanation:
For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
A₁ v₁ = A₂ v₂
Let's look for the areas
r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm
r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm
A₁ = π r₁²
A₁ = π 1.125²
A₁ = 3,976 cm²
A₂ = π r₂²
A₂ = π 0.1²
A₂ = 0.0452 cm²
Now with the continuity equation we can look for the speed of water inside the hose
v₁ = v₂ A₂ / A₁
v₁ = 11.2 0.0452 / 3.976
v₁ = 0.1273 m / s
Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)
P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂
P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂
Let's calculate
P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25
P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴
P₁ = 2.3506 10⁵ Pa