Answer:
F = 0.1153 N
Explanation:
Given:
- Rain fall rate = 5 cm/hr
- Velocity before impact vi = 8.3 m/s
- The Area of the pan A = 1.0 m^2
- The density of water ρ = 1000 kg/m^3
Find:
Estimate the magnitude of the force
Solution:
- Consider rain drops impacting a surface looses all its momentum. So the change in its momentum is just the momentum with which it impacted.
- We do not know the size of each drop. But the rain fall rate allows us to calculate the rate of change of momentum.
- Total Force: The total force experience by the surface due to the momentum transfer from the impacting rain drops is:
F = m(Δv) / t = ρ*v*A*Δh/Δt
Where
Δh/Δt = rain-fall rate.
F = 1000*8.3*1*0.05 / 3600
F = 0.1153 N
The magnitude of the force is 0.1153N
Given-
The velocity of the raindrop=8.3m/sec
Let when the raindrop hits the surface it loses its momentum
Now the magnitude of the force on the bottom of the given area can be calculated as
[tex]F=\dfrac{m\bigtriangleup v}{t}[/tex]
[tex]F= \rho vA \frac{\bigtriangleup h}{\bigtriangleup t}[/tex]
here rainfall rate
=[tex]\dfrac{\bigtriangleup h}{\bigtriangleup t}[/tex]
therefore,
[tex]F=1000\times 8.3\times 1 \times \dfrac{0.05}{3600}[/tex]
[tex]F=0.1153N[/tex]
Hence the magnitude of the force is 0.1153N
For more detail about the momentum, follo9w the link
https://brainly.com/question/4956182
