To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,
M,m = Counts per second
Our radios are given by
Therefore replacing we have that,
Therefore the number of counts expect at a distance of 20 cm is 19.66cps
Average force on the window: 0.32 N
Explanation:
The average force exerted on the window is given by Newton's second law
where
is the net change in momentum of the hailstones in a time interval of
In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:
is the initial momentum of one hailstone, with
m = 7 g = 0.007 kg is the mass
is the initial speed
is the angle with the window
The final momentum is
where
v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)
(after the rebound, the direction has changed)
So the change in momentum of 1 hailstone is
We are interested only in the magnitude, so
There are 595 hailstones hitting the window in 49 s, so the total change in momentum is
And therefore, the average force on the window is
Learn more about force:
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