help please!!!?

Question

An Olympic-size pool is 40. m long and 30. m wide.
(a) How many gallons of water (d = 1.0 g/mL) are needed to fill the pool to an average depth of 6.0 ft?
(b)What is the mass (in kg) of water in the pool?

Answer 1

Note - to convert units, you can go to google and type, for instance: 2194.56m^3=?ml

--Volume of the Pool--
6ft = 1.8288m

therefore the volume of the pool:
V = 1.8288m * 30m * 40m
= 2194.56m^3

(a)--Volume of Water in Gallons--
We know that it would take 2194.56m^3 of water to fill the pool.

2194.56m^3 = 579 741.417 US gallons

(b)--Mass of the Water--
The volume of the pool is 2194.56m^3

2194.56m^3 = 2 194 560 000 ml

if the density is 1.0 g/mL, then:

d = m/v
therefore m = dv
m = (1g.ml)(2 194 560 000 ml)
m = 2 194 560 000 g * (1kg/1000g)
m = 2 194 560 kg

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Answer 2

First of all you have 2 systems here (English and metric). You don't want that. For the first part, you need to convert everything to English units. There are 39.3700787 inches in a meter and 72 inches in 6 feet. We then get the total number of inches of our dimentions.
40 * 39.3700787 = 1574.803148 inches
30 * 39.3700787 = 1181.102361 inches
6 ft * 12 inchess = 72 inches
Now we multiply these togather giving us:
133920267.567338334816 inches ^ 3
and there are 231 in^3 / gallon so we divide by 231:
and get 579741.41804042569184415584415584 Gallons
or 579,741.418 Gallons.

b)
multiply this by 3.8 liter/gallon
2203017.3884 liters
We know that water's density is 1 g/ml and there 1000 g / Kg and 1000 ml / liter
so we have about (for rounding errors) 2, 203, 017.3884 kilograms of water in the pool.

Answer 3

First convert 6 feet to meters

1 foot = 0.3048 meters

Multiply 6 times 0.3048

6 x 0.3048 = 1.8288 Meters
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Multiply 40 times 30 times 1.8288

40 x 30 x 1.8288 =

1200 x 1.8288 =

2194.56 Cubic meters

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1 cubic meter of water is 264.1720523578 gallons

Multiply 2194.56 times 264.1720523578 to get total gallons

2194.56 x 264.1720523578 =

579741.4192223942 gallons

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