A 515 cm3 block of aluminum (Density = 2.7 g/cm3) is lowered carefully into a completely full beaker of water. What is the weight of the water, in newtons, that spills out of the beaker?
assuming that the beaker is of a size and shape to completely contain the block, an amount of water equal to the block's volume will be displaced.
515cm^3 water = 515 g = .515kg
weight = 9.81 * .515
To find the weight of the water that spills out of the beaker, we need to determine the volume of the water displaced by the aluminum block and then calculate its weight.
The volume of the aluminum block is given as 515 cm³. The density of aluminum is also given as 2.7 g/cm³. Therefore, the mass of the aluminum block can be calculated as follows:
Mass of aluminum block = Volume of aluminum block × Density of aluminum
Convert the density of aluminum from grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³) for consistency:
Density of aluminum = 2.7 g/cm³ = 2700 kg/m³
Now, we can calculate the mass of the aluminum block:
Mass of aluminum block = 515 cm³ × 2700 kg/m³
Next, we need to calculate the weight of the aluminum block using the formula:
Weight = Mass × gravitational acceleration
The gravitational acceleration is approximately 9.8 m/s².
Weight of aluminum block = Mass of aluminum block × 9.8 m/s²
Now, let's find the weight of the water that spills out of the beaker. Since the beaker was completely full, the weight of the water spilled equals the weight of the aluminum block.
Therefore, the weight of the water that spills out of the beaker is equal to the weight of the aluminum block, which we calculated earlier.