004) The force F exerted by each hailstone is given by Second Newton law:
[tex]F=ma[/tex]
where m is the mass of each hailstone and a is its acceleration. a can be calculated by using:
[tex]a=\frac{\Delta v}{t}[/tex]
where v is the speed at which the hailstones impact the window and t the time interval of the impact.
By replacing the values of v and t you obtain for acceleration:
[tex]a=\frac{4.1\frac{m}{s}}{29s}=0.1413793103\frac{m}{s^2}[/tex]
Then, the force F is:
[tex]F=(0.004kg)(0.1413793103\frac{m}{s^2})=5.65517241\times10^{-4}N[/tex]
where the mass m is used in kg in order to obtain N units for force.
Now, consider that the impact of the hailstones on the window is at angle of 44 degrees. It means that the effective force exerted by each hailstones is:
[tex]55.65517241\times10^{-4}N\cdot\sin 44=3.92841285\times10^{-4}N[/tex]
It means that the average force due to the total number of hailstones is:
[tex]F_{\text{avg}}=2\cdot3.92841285\times10^{-4}N\cdot525=0.4124833496N[/tex]
005) Then, by replacing the previous value of the force and the area of the impact on the window, into the formula for the pressure, you have:
[tex]P=\frac{0.4124833496^{}N}{0.974m^2}=0.4234941988\frac{N}{m^2}[/tex]
