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The only swimming pool at the El Cheapo Motel is outdoors. It is 5.0 m wide and 12.0 m long. If the weekly evaporation is 2.35 in., how many gallons of water must be added to the pool if it does not rain?

Answers

Answer 1
Answer:

Answer:

946.10 gallons per week

Step-by-step explanation:

1 cm = 0.393701 inch

Width = 5.0m = 196.85 inch

Length = 12.0 m = 472.44 inch

The volume evaporated weekly is given by:

V = L*W*2.35 = 196.85*472.33*2.35\nV=218,550.12\ in^3

Converting to gallons:

1\ gal = 231\ in^3\nV=(218,550.12)/(231)\nV= 946.10\ gal

946.10 gallons of water must be added to the pool each week.
Answer 2
Answer:

Final answer:

To determine the volume of water evaporated from the pool at the El Cheapo Motel, we converted all measurements to a common unit and calculated the volume of water evaporated. The motel has to add approximately 946.13 gallons of water weekly, considering there is no rain.

Explanation:

To answer this question, we first need to convert the measurements to a common unit. Given that the pool is 5.0 m wide and 12.0 m long (a total area of 60.0 m2) and the weekly evaporation is 2.35 inches, we first convert the inches to meters. Since 1 inch is equal to 0.0254 meters, 2.35 inches equals 0.05969 meters.

Then, we calculate the volume of water evaporated in a week, which is calculated by multiplying the surface area of the pool by the depth of the water evaporated. Hence, it's 60.0 m2 * 0.05969 m = 3.58 m3. As 1 m3 is approximately 264.17 gallons, 3.58 m3 equals 946.1296 gallons approximately.

In conclusion, the El Cheapo Motel needs to add around 946.13 gallons of water to their pool on a weekly basis, if there is no rain.

Learn more about Volume Calculation here:

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