On September 3, 1970 a hailstone with diameter 5.6in fell at Coffeyville, Kansas. It weighed about 0.018Ib/in^3 compared to the normal 0.033Ib/in^3 for ice. About how heavy was this Kansas hailstone?
1.7 lb
4/3 pi (5.6/2)^3 in^3 * 0.018 lb/in^3 = ? lbs
the nonsense about "normal" density of ice is just noise.
To find the weight of the hailstone, we can use the formula:
Weight = Volume x Density
First, we need to calculate the volume of the hailstone using the diameter.
The formula for the volume of a sphere is: V = (4/3) x π x r^3
The radius (r) of the hailstone can be calculated by dividing the diameter by 2:
r = 5.6in / 2 = 2.8in
Now we can calculate the volume of the hailstone:
V = (4/3) x π x (2.8in)^3
V ≈ 4.18879 x (2.8in)^3
V ≈ 4.18879 x 21.952in^3
V ≈ 91.94in^3
Next, we can calculate the weight using the volume and the reduced density:
Weight = 91.94in^3 x 0.018Ib/in^3
Weight ≈ 1.655 Ibs
Therefore, the estimated weight of the hailstone that fell in Coffeyville, Kansas on September 3, 1970, was approximately 1.655 pounds.
To find the weight of the Kansas hailstone, we can use the formula:
Weight = Volume x Density
First, let's calculate the volume of the hailstone. The formula for the volume of a sphere is given by:
Volume = (4/3) x π x (radius)^3
We know the diameter of the hailstone is 5.6 inches, so the radius would be half of that:
radius = diameter / 2 = 5.6 inches / 2 = 2.8 inches
Now we can calculate the volume:
Volume = (4/3) x π x (2.8 inches)^3 ≈ 91.77 cubic inches
Next, we can find the weight of the hailstone using the given density:
Weight = Volume x Density
Weight = 91.77 cubic inches x 0.018 lb/in^3 ≈ 1.65 pounds
Therefore, the Kansas hailstone would be approximately 1.65 pounds heavy.