A cylindrical tank lying on its side is filled with liquid weighing 50lb/ft^3. Find the work required to pump all the liquid to
a level 1 ft above the top of the tank.The diameter of the tank is 4 feet.The depth of the tank is 10 feet. NOTE: Please remember the tank is lying on its SIDE, it's NOT upright. A hint my professor gave is that the answer should be a number multiplied by pi. I'm not really sure where my slice goes and how to set up my integral because it's not upright.
<span>You can use the fact that the work done is equivalent to finding the center of mass of the water in the tank(which is at the center of the cylindrical cross section, which is 3 ft below the level to which the liquid is pumped. Call this distance h. You then need the total mass of the liquid, volume X density=m. This is where pi comes in! Work needed =mgh, where g is the gravitational acceleration.</span><span> </span>
Step-by-step explanation: First, turn 15% into the decimal 0.15, then multiply by 45, which equals 6.75. Now add 6.75, which is the amount the waitress got tipped to 45, which equals 51.75$ total.
