The mean pressure is `(6K)/5`, which rain renders to vertical windshield of automobile, moving with constant velocity of magnitude `v=12m//s`. Consider that raindros fall vertically with speed `u=5m//s`. The intensity of rainfall deposits `h=2cm` of sediments in time `tau=1` minute: [ `rho=10^(3)kg//m^(3))` is the density of liquid] (Assume collisions are inelastic). Calculate `K`.

Explanation

To solve the problem, we need to calculate the value of \( K \) given the mean pressure exerted by the rain on the windshield of an automobile. Here’s the step-by-step solution:

### Step 1: Understanding the Problem
We know that the mean pressure \( P \) is given by:
\[ P = \frac{6K}{5} \]
The automobile is moving with a constant velocity \( v = 12 \, \text{m/s} \) and the raindrops fall vertically with a speed \( u = 5 \, \text{m/s} \). The intensity of rainfall deposits \( h = 2 \, \text{cm} = 0.02 \, \text{m} \) of sediments in time \( \tau = 1 \, \text{minute} = 60 \, \text{s} \). The density of the liquid is \( \rho = 10^3 \, \text{kg/m}^3 \).

### Step 2: Calculate the Mass of Rain Deposited
The mass of rain deposited on the windshield can be calculated using the volume of rain that falls and the density of the rain. The volume of rain that falls in time \( \tau \) over an area \( S_0 \) is:
\[ \text{Volume} = h \cdot S_0 = 0.02 \cdot S_0 \, \text{m}^3 \]
The mass of the rain is:
\[ \text{Mass} = \text{Volume} \cdot \rho = 0.02 S_0 \cdot 10^3 = 20 S_0 \, \text{kg} \]

### Step 3: Calculate the Number Density of Raindrops
The number density \( \lambda \) of raindrops can be expressed as:
\[ \lambda = \frac{\text{Mass}}{\tau \cdot S_0 \cdot u} \]
Substituting the values we have:
\[ \lambda = \frac{20 S_0}{60 \cdot S_0 \cdot 5} = \frac{20}{300} = \frac{1}{15} \, \text{kg/m}^3 \]

### Step 4: Calculate the Force Exerted by the Rain
The force \( F \) exerted by the rain on the windshield can be calculated using the change in momentum due to the inelastic collision. The force can be expressed as:
\[ F = \lambda \cdot S \cdot v^2 \]
Where \( S \) is the area of the windshield. Thus:
\[ F = S \cdot v^2 \cdot \lambda \]
Substituting \( \lambda \):
\[ F = S \cdot v^2 \cdot \frac{1}{15} = S \cdot 12^2 \cdot \frac{1}{15} = S \cdot \frac{144}{15} = S \cdot 9.6 \, \text{N} \]

### Step 5: Calculate the Pressure
Pressure \( P \) is defined as force per unit area:
\[ P = \frac{F}{S} = \frac{S \cdot 9.6}{S} = 9.6 \, \text{Pa} \]

### Step 6: Equate to Find \( K \)
Now, we equate this pressure to the given mean pressure:
\[ 9.6 = \frac{6K}{5} \]
Multiplying both sides by 5:
\[ 48 = 6K \]
Dividing both sides by 6:
\[ K = 8 \]

### Final Answer
Thus, the value of \( K \) is:
\[ \boxed{8} \]