Question

The great limestone caverns were formed by dripping water. The acceleration of gravity is 9.81 m/s 2 . If water droplets of 0.26 mL fall from a height of 9 m at a rate of 15 per minute, what is the average force exerted on the limestone floor by the droplets of water during a 7 min period? Answer in units of N.

Answer 1

Step-by-step explanation:

In order to calculate the average force exerted we need to calculate the mass of water droplet first of all as follows.

Volume =
`0.26 ml * (1 * 10^(-6) m^(3))/(1 mL)`

=
`2.6 * 10^(-7) m^(3)`

As we know that density of water is 1000
`kg/m^(3)`. The mass is given as follows.

m =
`\rho V`

=
`1000 kg/m^(3) * 2.6 * 10^(-7) m^(3)`

=
`2.6 * 10^(-4) kg`

As the given water droplet falls from a height of 9 m. Therefore, its velocity when it hits the ground is as follows.

v =
`√(2gh)`

=
`\sqrt{2 * 9.8 m/s^(2) * 9}` m/sec

= 13.28 m/sec

Now, we will calculate the rate of droplet fall as follows.

R =
`(15)/(1 min) * (1 min)/(60 sec)`

= 0.25 drops/sec

Therefore, force will be calculated as follows.

F =
`R (\Delta p)/(\Delta t)` (
`\Delta p = m * v`)

=
`0.25 drops/sec * (2.6 * 10^(-4) kg * 13.28 m/sec)/(1)`

= 8.632 N

Thus, we can conclude that 8.632 N is the average force exerted on the limestone floor by the droplets of water during a 7 min period.