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anonymous
 5 years ago
In a 30.0s interval, 500 hailstones strike a glass window
with an area of 0.600m2 at an angle of 45.0°to the window surface. Each hailstone has a mass of 5.00g and a
speed of 8.00 m/s. If the collisions are elastic, what are
the average force and pressure on the window?
anonymous
 5 years ago
In a 30.0s interval, 500 hailstones strike a glass window with an area of 0.600m2 at an angle of 45.0°to the window surface. Each hailstone has a mass of 5.00g and a speed of 8.00 m/s. If the collisions are elastic, what are the average force and pressure on the window?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f = ma. m = 0.005 kg, a = 9.8 m/s^2. f = ma sin(theta) = 0.005*9.8*sin(45) = 0.005*9.8*1/2 = 1/400 * 9.8 = 0.0245 all in my head

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For one hailstone we have; Force = Mass X acceleration = 0.005kg x 9.8.} This is when the hailstone is not inclined at an angle. When the hailstone is inclined at an angle of 45, then the component of force along the glass window will be F =0.005kg x 9.8 x sin45= 0.005kg x 9.8 x 0.707= 0.0346N. Therefore, total force for the 500 hailstones would be 500x0.0346N=17.32N This force is acting on an area equal to 0.600m2 Pressure = Force per unit area = 17.32N/0.600m2 = 28.9Pa
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