The level of water in an olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2.00 meters deep) needs to be lowered 3.00 cm. if water is pumped out at a rate of 3.80 liters per second, how long will it take to lower the water level 3.00 cm?

Initial dimension of the pool:

Length (L)= 50.0 m

Width (W)= 25.0 m

Depth (D) = 2.0 m

Initial Volume of the pool (V1)  = L*W*D = 50.0 * 25.0 *2.0 = 2500 m3

When the depth is lowered by 3 cm i.e. 0.03 m, the new depth becomes: 2.0 - 0.03 = 1.97 m

The new volume of the pool (V2) = 50.0*25.0*1.97 = 2462.5 m3

Volume of water to be pumped out = V1-V2 = 2500-2462.5 = 35.5 m3

Now, 1 m3 = 1000 L

therefore, 35.5 m3 == 35.5 *10^3 L

It is given that:

3.80 L of water is pumped out in 1 sec

Therefore, 35.5*10^3 L will be pumped out in:

= 1 s * 35.5*10^3 L/3.80 L = 9.34*10^3 sec

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

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