Volumes of regular objects like the aluminum bar and PE (rod) were calculated conventionally using V = lwh. The irregular iron bolt's volume was determined through water displacement, addressing its shape complexities.
In Procedure step 2, the volumes of the samples were calculated using the formula V = lwh (volume equals length times width times height) for regular objects. For the aluminum bar, with dimensions 9.7 cm (length), 1.6 cm (width), and 1.0 cm (height), the calculated volume is 15.52 cm³. Similarly, for the PE (rod) with a diameter of 1.7 cm and a length of 10.5 cm, the calculated volume is 23.84 cm³.
However, the iron bolt's irregular shape makes it challenging to use the same formula. Without provided dimensions, the volume cannot be calculated conventionally. Instead, the displacement volume method was employed. By subtracting the initial water volume from the volume after immersing the bolt (6.95 cm³), the actual volume is determined.
The irregularity of the iron bolt, lacking well-defined geometric dimensions, hinders volume calculation through the regular formula. The water displacement method overcomes this limitation by indirectly measuring the volume. In summary, while the volumes of the aluminum bar and PE (rod) are calculated conventionally, the irregular shape of the iron bolt necessitates an alternative method like water displacement.
The question probable may be:
Dry Lab 2 - Questions for Mass, Volume, and Density MaterialMass, gDimensions,cmCalculatedVolume,cm^3DisplacementVolume,cm^3 Density, g/mL Aluminum(bar)39Length: 9.7Width: 1.6 Height: 1.0 15.5240-25=152.52 PE (rod)21Diameter: 1.7Length: 10.5 23.848-25=230.88 Iron (bolt)39N/AN/A31.9-25=6.95.65 1.Calculate the volumes of the samples used in Procedure step 2. Record the calculated volumes in Table 2.2. Why could we not find the volume of the bolt by this method?