Question

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m^3 , length 92.4 cm and diameter 2.15 cm from a storage room to a machinist. Calculate the weight of the rod, w. Assume the free-fall acceleration is g =9.81m/s^2

Answer 1

Answer:

25.68 N

Explanation:

From the question given above, the following data were obtained:

Density of cylindrical rod = 7800 kg/m³

Length = 92.4 cm

Diameter = 2.15 cm

Acceleration due to gravity (g) = 9.8 m/s²

Weight of rod =?

Next, we shall determine the volume of the rod. This can be obtained as follow:

Height (h) = 92.4 cm

Diameter (d) = 2.15 cm

Pi (π) = 3.14

Volume (V) =?

V = π(d/2)²h

V = 3.14 × (2.15/2)² × 92.4

V = 335.29 cm³

Next, we shall convert 335.29 cm³ to m³. This can be obtained as follow:

1 cm³ = 1×10¯⁶ m³

Therefore,

335.29 cm³ = 335.29 cm³ × 1×10¯⁶ m³ / 1 cm³

335.29 cm³ = 0.00033529 m³

Thus, 335.29 cm³ is equivalent to 0.00033529 m³.

Next, we shall determine the mass of the rod. This can be obtained as follow:

Density of rod = 7800 kg/m³

Volume of rod = 0.00033529 m³.

Mass of rod =?

Density = mass /volume

7800 = mass / 0.00033529

Cross multiply

Mass of rod = 7800 × 0.00033529

Mass of rod = 2.62 Kg

Finally, we shall determine the weight of the rod. This can be obtained as follow:

Mass (m) of rod = 2.62 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) of rod =?

W = m × g

W = 2.62 × 9.8

W = 25.68 N

Therefore, the weight of the rod is 25.68 N