Homework SourseA circular swimming pool has a diameter ofm The sides arem h
A circular swimming pool has a diameter of![\"18\"](\"//mathweb1.sandbox.csun.edu/webwork2_files/tmp/equations/60/395773d8ab9cf8c969d8c71b56429b1.png\") m. The sides are![\"3\"](\"//mathweb1.sandbox.csun.edu/webwork2_files/tmp/equations/e6/5935e933d7cde7e57dfe74118766791.png\") m high and the depth of the water is![\"2.5\"](\"//mathweb1.sandbox.csun.edu/webwork2_files/tmp/equations/62/f091b82b351c72e50674fe5870d49d1.png\") m. How much work (in Joules) is required to pump all of the water over the side?(The acceleration due to gravity is ![\"9.8](\"//mathweb1.sandbox.csun.edu/webwork2_files/tmp/equations/5f/921f87245a574fc6240b82283d426f1.png\") and the density of water is![\"1000](\"//mathweb1.sandbox.csun.edu/webwork2_files/tmp/equations/c0/db5989c0ef9ab45aead021677d9df01.png\") .)Answer:Joules |
Solution
Volume of water = r2h = *9*9*2.5 = 636.172 m3
Hence mass of water = V* = 636172 kg
Now centre of mass of water when it is inside pool is 2.5/2 = 1.25 mtrs above base.
And to remove all water from the pool this has to be raised by 3 - 1.25 mtrs = 1.75 mtrs
Hence work to be done = mgh = 636172*9.8*1.75 = 10910349.8 J
The answer may be off by around ±10 joules due to rounding in calculating volume
