Answer:
1.265 Pounds
Explanation:
Data provided:
Tire outside diameter = 49"
Rim diameter = 22"
Tire width = 19"
Now,
1" = 0.0254 m
thus,
Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m
Rim radius, r₂ = 22" / 2 = 11" = 0.2794 m
Tire width, d = 19" = 0.4826 m
Now,
Volume of the tire = π ( r₁² - r₂² ) × d
on substituting the values, we get
Volume of air in the tire = π ( 0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³
Also,
Density of air = 1.225 kg/m³
thus,
weight of the air in the tire = Density of air × Volume air in the tire
or
weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg
also,
1 kg = 2.204 pounds
Hence,
0.5742 kg = 0.5742 × 2.204 = 1.265 Pounds
