Answer:
t = 2.26 10⁵ s
Explanation:
Let's calculate the flow of water that comes out of the hose
Q = A v
The area of a circle is
A = π r²
Q = π r² v
Let's reduce the units to the SI system
d = 5/8 in (2.54 10⁻² m / in) = 1.5875 10⁻² m
r = d / 2 = 0.79375 10⁻² m
Q = π (0.79375 10⁻²)² 0.59
Q = 1.1678 10⁻⁴ m³ / s
Let's calculate the volume of the pool
V = π R² h
V = π (4.6 / 2)² 1.6
V = 26.59 m³
Let's use a rule of proportions (rule of three), to find the time
t = 26.59 (1 / 1.1678 10⁻⁴)
t = 22.77 10⁴ s
t = 2.26 10⁵ s
Answer:
Time taken to fill the pool(t) = 223529.41 s
Explanation:
The volume of water in the pool (V) = πd²h/4............... equation 1
Where d = radius of the pool, h = height of the pool
At a height of 1.6 m, the volume of water is
V(pool) = (3.143 × 4.6² × 1.6)/4 = 26.60 m³.
The volume of the hose is
V(hose) = πd²h/4........................... equation 2
And the rate of flow of water from the hose into the pool is
V(hose)/dt = (dh/dt) × (dV(hose)/dh)
Differentiating equation 2,
V(hose)/dt = πd²/4
where d = 5/8-in. =(5/8)×0.0254 =0.016 m, π = 3.143, dh/dt = 0.59 m/s
dV(hose)/dt = {(3.143×0.016²)/4} × 0.59
dV(hose)/dt = 0.000119 m³/s.
∴ time taken to fill the pool(t) = V(pool)/(dV(hose)/dt)
t = 26.60/0.000119
t = 223529.41 s.
Time taken to fill the pool(t) = 223529.41 s
