Answer:
The work required to pump all of the water over the side is 3.991×10^7 Joules.
Explanation:
Data given:
Diameter (D) = 18 m
Height = 4 m
Depth = 4 m
Work required to pump all of the water over the side is calculated by multiplying the pressure in the pool by the volume of the pool
Pressure (P) = height of swimming pool × density of water × acceleration due to gravity = 4 m × 1000 kg/m^3 × 9.8 m/s^2 = 39,200 kg/m.s^2 = 39,200 N/m^2
Volume (V) = area of pool × depth = πD^2/4 × depth = 3.142×18^2/4 × 4 = 1018.008 m^3
Work required = PV = 39,200 N/m^2 × 1018.008 m^3 = 3.991×10^7 Nm = 3.991×10^7 J
Answer:
The work required to pump all of the water over the side is 2.0 x 10⁷ J
Explanation:
Given;
diameter of the circular pool, D = 18 m
radius of the pool, r = 18/2 = 9m
depth of the pool, h = 4 m
depth of the side, H = 4 m
Work done = force x distance
Force of the water in the pool = ρVg
where;
ρ is the density of water = 1000 kg/m³
V is the volume of the pool
g is the acceleration due to gravity
Volume of the pool = πr²dy = π(9)²dy
Volume of the pool = 81π dy
Force = ρVg
Force = (1000) x (81πdy) x (9.8) = 793800π dy
Let the distance in which the water is moved = y
Work done = Fd = Fy
Work done = (793800π dy) x (y)
Work done = 793800πy dy
[tex]Work\ done = \int\limits^4_0 {793800\pi y} \, dy\\\\Work\ done = 793800\pi \int\limits^4_0 {y} \, dy \\\\Work \ done = 793800\pi [\frac{y^2}{2}]^4_0\\\\Work \ done = \frac{793800\pi}{2} [y^2]^4_0\\\\Work \ done = 396900\pi (4^2-0)\\\\Work \ done = 396900\pi *16\\\\Work \ done = 6350400\pi \ J\\\\Work \ done = 19952956.8 \ J[/tex]
Work done = 2.0 x 10⁷ J
Therefore, the work required to pump all of the water over the side is 2.0 x 10⁷ J
