The weight of the cylindrical iron rod of density 7800 kg/m³ , length 84.8 cm and diameter 2.50 cm is 31.8 N
The relation between mass and density is given by:
m = ρ . V
Where:
m = mass
ρ = density
V = volume
In the given problem.
ρ = 7800 kg/m³
length (l) = 84.8 cm
diameter = 2.5 cm,
radius (r) = 2.5/2 = 1.25 cm
Hence, the volume of the rod is:
V = πr² . l
= 3.14 x 1.25² x 84.8 = 416 cm³
= 416 x 10⁻⁶ m³
Hence, its mass is:
m = 7800 x 416 x 10⁻⁶ = 3.24 kg
And its weight is:
w = m x g
= 3.24 x 9.8 = 31.8 N
Hence, the weight is 31.8 N
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