What is the apparent weight of a lead brick 2.0 in x 2.0 in x 8.0 in if placed in oil with density p=0.93 g/cm^3 (Ppb=11.4 g/cm^3)(1 in. = 2.54 cm)

5.5kg

To calculate the apparent weight of the lead brick when placed in oil, we can use Archimedes' principle. According to Archimedes' principle, the apparent weight of an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

Here's how you can calculate the apparent weight of the lead brick:

Step 1: Convert the dimensions of the lead brick from inches to centimeters.
The given dimensions of the lead brick are 2.0 in x 2.0 in x 8.0 in. Since 1 inch is equal to 2.54 centimeters, we can convert the dimensions as follows:
Length = 2.0 in x 2.54 cm/in = 5.08 cm
Width = 2.0 in x 2.54 cm/in = 5.08 cm
Height = 8.0 in x 2.54 cm/in = 20.32 cm

Step 2: Calculate the volume of the lead brick.
The volume of the lead brick can be calculated by multiplying its length, width, and height:
Volume = Length x Width x Height = 5.08 cm x 5.08 cm x 20.32 cm = 520.181888 cm^3 (rounded to three decimal places)

Step 3: Calculate the weight of the oil displaced by the lead brick.
The weight of the oil can be calculated using its density and the volume of the lead brick:
Weight of oil = Density x Volume = 0.93 g/cm^3 x 520.181888 cm^3 = 483.354478 g (rounded to three decimal places)

Step 4: Calculate the apparent weight of the lead brick.
The apparent weight of the lead brick is equal to the weight of the oil displaced by it:
Apparent weight = Weight of oil = 483.354478 g (rounded to three decimal places)

Therefore, the apparent weight of the lead brick when placed in oil with a density of p=0.93 g/cm^3 is approximately 483.354 g.