A pool is 41.4 m long and 28.2 m wide. If the average depth of water is 5.00 ft, what is the mass (in kg) of water in the pool? Enter your answer in scientific notation. The den…

Question

asked Sep 17, 2021 58.9k views

1 Answer

Sergi And Replace · Answer 1 · 2021-09-20T15:23:36+0000

Answer:

m=1.779x10^(6)kg

Step-by-step explanation:

Hello,

In this case, mass and volume define density as shown below:

$\rho =(m)/(V)$

Whereas the mass is computed by:

$m=\rho *V$

Thus, knowing the dimensions of the pool, we compute its volume:

V=41.4m*28.2m*(5.00ft*(1m)/(3.281ft) )=1779.15m^3

But in cubic centimetres:

$V=1779.15m^3*(1x10^6cm)/(1m^3) \\\\V=1.779x10^(9)cm^3$

Therefore, the mass in kilograms turns out:

$m=1(g)/(cm^3) *1.779x10^(9)cm^3*(1kg)/(1000g)\\ \\m=1.779x10^(6)kg$

Regards.