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

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.