Respuesta :
Average force on the window: 0.32 N
Explanation:
The average force exerted on the window is given by Newton's second law
[tex]F=\frac{\Delta p}{\Delta t}[/tex]
where
[tex]\Delta p[/tex] is the net change in momentum of the hailstones in a time interval of [tex]\Delta t[/tex]
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:
[tex]p_i =m u sin \theta[/tex] is the initial momentum of one hailstone, with
m = 7 g = 0.007 kg is the mass
[tex]u=4.5 m/s[/tex] is the initial speed
[tex]\theta=25^{\circ}[/tex] is the angle with the window
The final momentum is
[tex]p_f = mv sin \theta[/tex]
where
v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)
[tex]\theta=-25^{\circ}[/tex] (after the rebound, the direction has changed)
So the change in momentum of 1 hailstone is
[tex]\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s[/tex]
We are interested only in the magnitude, so
[tex]\Delta p = 0.0266 kg m/s[/tex]
There are 595 hailstones hitting the window in 49 s, so the total change in momentum is
[tex]\Delta p = 595\cdot 0.0266 = 15.8 kg m/s[/tex]
And therefore, the average force on the window is
[tex]F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N[/tex]
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The average force on the window is 0.323N
Elastic collision:
Since the collision is elasctic the hailstone bounces back with the same velocity.
Therefore the change in momentum of the hailstone is:
Δp = 2 × mass × velocity
But the hailstone is inclined at an angle 25° so
For one hailstone we have;
Δp = 2×0.007×4.5×sin25
Δp = 0.0266 kgm/s
For 595 hailstones
ΔP = 595 × 0.0266
ΔP = 15.82 kgm/s
Now the average force is the rate of change of momentum:
F = ΔP/Δt
F = 15.82/49
F = 0.323N
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