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.

The value of mass in kg is .

Length , l = 59.8 m.

Breadth , b  = 26.6 m.

Depth , d = 3.7 ft  

Volume of pool is as follows:-

It is defined as mass per unit volume.

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

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Given :

Length , l = 59.8 m.

Breadth , b  = 26.6 m.

Depth , d = 3.7 ft .

Density of water , .

To Find :

Mass of water in pool .

Solution :

First we will covert depth into m from ft .

For ,

So , volume of pool is :

We know , density is given by :

So ,

Putting given values in above equation :

Hence , this is the required solution.

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j