Final answer:
The problem asks for the work required to pump water to the top of a cylindrical tank, involving the calculation of force due to gravity on the water's mass and the distance it must be moved, then converting the energy to mega-joules.
Step-by-step explanation:
The question entails a physics problem related to work done and mechanical energy. To find the work done in pumping water to the top of a cylindrical tank, we need to use the formula for work: W = F × d, where F is the force and d is the distance. The force in this context will be the weight of the water, which is calculated by mass (m) times gravity (g). The mass of the water can be found by multiplying the volume (V) of the half-full tank with the density (ρ) of water.
Since we're given the radius (r) of the tank is 4 meters, and the tank is half full of an 18-meter tall tank, the volume of water in the tank is V = πr^2h, where h is 9 meters (half of 18 meters). The density of water is 1000 kg/m³, and the constant for gravity is 9.8 m/s². After calculating the mass, we need to consider that the water has to be lifted an average height from its current position to the top of the tank, which will be less than 9 meters since the water level reduces as it gets pumped out.
To simplify calculations, we consider the average height h_{avg} which is the average distance a mass element needs to be lifted to reach the top. The work done to pump out the entire volume of water is given by multiplying the total mass of water m by the average height h_{avg} and gravity g, giving W = mgh_{avg}. Finally, to express this energy in mega-joules (MJ), we need to divide the result by 1 million (10^6), since 1 joule is 10^-6 mega-joules.