Respuesta :
Explanation:
In order to calculate the average force exerted we need to calculate the mass of water droplet first of all as follows.
Volume = [tex]0.26 ml \times \frac{1 \times 10^{-6} m^{3}}{1 mL}[/tex]
= [tex]2.6 \times 10^{-7} m^{3}[/tex]
As we know that density of water is 1000 [tex]kg/m^{3}[/tex]. The mass is given as follows.
m = [tex]\rho V[/tex]
= [tex]1000 kg/m^{3} \times 2.6 \times 10^{-7} m^{3}[/tex]
= [tex]2.6 \times 10^{-4} kg[/tex]
As the given water droplet falls from a height of 9 m. Therefore, its velocity when it hits the ground is as follows.
v = [tex]\sqrt{2gh}[/tex]
= [tex]\sqrt{2 \times 9.8 m/s^{2} \times 9}[/tex] m/sec
= 13.28 m/sec
Now, we will calculate the rate of droplet fall as follows.
R = [tex]\frac{15}{1 min} \times \frac{1 min}{60 sec}[/tex]
= 0.25 drops/sec
Therefore, force will be calculated as follows.
F = [tex]R \frac{\Delta p}{\Delta t}[/tex] ([tex]\Delta p = m \times v[/tex])
= [tex]0.25 drops/sec \times \frac{2.6 \times 10^{-4} kg \times 13.28 m/sec}{1}[/tex]
= 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.
