Title: work Post by: on February 06, 2004, 03:29:00 am can someone please please please help me with this? I am a lost little confused college student close to dropping out. please help!!!!!
A rectangulat tank that is three feet long and two feet wide and two feet deep contains water that is 1.5 feet in depth calculate the amount of work required to pump half of the water over the top of the tank (weight density of water is 62.5 pounds per cubic foot)? Title: topic88 Post by: FuKwun Hwang on February 06, 2004, 04:41:26 pm Use conservation of energy:
Calculate the total potential energy (related to water surface) for upper half of water ! $U= \int dm \cdot g \cdot y =\int \rho \cdot dV \cdot g \cdot y =\int \rho \cdot A \cdot dy \cdot g \cdot y $ where $\rho$ is the density, A is the area Title: Re: work Post by: adrianbalaba on June 29, 2007, 01:39:38 pm A little bit of confused... What is that?
