Question

A pool is 59.8 m long and 26.6 m wide. If the average depth of water is 3.70 ft, what is the mass (in kg) of water in the pool? Enter your answer in scientific notation. The density of water is 1.0 g/mL.

Answer 1

Given :

Length , l = 59.8 m.

Breadth , b  = 26.6 m.

Depth , d = 3.7 ft .

Density of water , $\rho=1\ g/ml=1000\ kg/m^3$ .

To Find :

Mass of water in pool .

Solution :

First we will covert depth into m from ft .

$1\ ft =0.3\ m$

For ,

$3.7\ ft=0.3\times 3.7\ m\\3.7\ ft=1.11\ m$

So , volume of pool is :

$V=59.8\times 26.6\times 1.11\ m^3\\\\V=1765.65\ m^3$

We know , density is given by :

$\rho=\dfrac{m}{V}$

So , $m=\rho V$

Putting given values in above equation :

$m=1000\times 1765.65\ kg\\\\m=1.77\times 10^6\ kg$

Hence , this is the required solution.

Answer 2

The value of mass in kg is $1.76\times10^{6} \ kg$.

Length , l = 59.8 m.

Breadth , b  = 26.6 m.

Depth , d = 3.7 ft  

Covert depth into m from ft.

$1ft=0.3m$

$3.7ft=0.3\times3.7\\=1.11m$

Volume of pool is as follows:-

$Volume=59.8\times26.6\times1.11\\=1765.65\ m^{3}\\=1765.65\times1000\ L=1765650 L$

Density:-

It is defined as mass per unit volume.

$Density=\frac{Mass}{Volume}$

Substitute 1.0 kg/L for density and 1765650 L for volume for the calculation of mass as follows:-

$m=1kg/L \times1765650\ L\\\\=1.76\times10^{6} \ kg$

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