The level of water in an Olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2 meters deep) needs to be lowered 3.2 cm. If the water is pumped out at a rate of 5.0 liters per second, how long will it take to lower the water level 3.2 cm?
volume of water to be removed is
5000 cm x 2500 cm x 3.2 cm = approx 4E7 cc
Convert to ? L
Then 5.0 L/sec x #sec = ? L.
Sole for #sec.
To find out how long it will take to lower the water level in the Olympic size swimming pool, we need to determine the volume of water that needs to be pumped out and then divide it by the rate at which the water is being pumped out.
First, let's calculate the volume of the water in the swimming pool. The pool's dimensions are given as 50.0 meters in length, 25.0 meters in width, and the water needs to be lowered by 3.2 centimeters (or 0.032 meters) in depth.
The formula to calculate the volume of a rectangular prism (in this case, the swimming pool) is:
Volume = length × width × depth
So, the volume of the swimming pool is:
Volume = 50.0 m × 25.0 m × 2.0 m
Volume = 2500 m³
Next, we need to convert the volume of the swimming pool from cubic meters to liters, since the pump rate is given in liters per second.
1 cubic meter (m³) = 1000 liters (L)
So, the volume of the swimming pool in liters is:
Volume = 2500 m³ × 1000 L/m³
Volume = 2,500,000 L
Now, we can calculate the time required to lower the water level by 3.2 cm (or 0.032 meters) at a rate of 5.0 liters per second.
Time = Volume / Pump Rate
Time = 2,500,000 L / (5.0 L/s)
Time = 500,000 seconds
Finally, we can convert the time from seconds to minutes or hours if desired. Since the question does not specify, we'll leave the time in seconds.
Therefore, it will take approximately 500,000 seconds to lower the water level in the Olympic size swimming pool by 3.2 centimeters.