hotonkylie147
09.12.2020 •
Physics
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
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Ответ:
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
Ответ:
Answer and Explanation:
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