Question

A 465 cm3 block of aluminum (D = 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?

Answer 1

To calculate the weight of the water that spills out of the beaker, we need to find the volume of the aluminum block and then use the density of water to calculate its weight. The weight of the water that spills out of the beaker is approximately 1.69 N.

Step-by-step explanation:

To calculate the weight of the water that spills out of the beaker, we need to find the volume of the aluminum block and then use the density of water to calculate its weight. The volume of the aluminum block can be found by dividing its mass by its density. In this case, the mass is given as 465 cm³ and the density of aluminum is 2.7 g/cm³. Therefore, the volume of the aluminum block is 465 cm³ / (2.7 g/cm³) = 172.22 g.

Now, we can use the volume of the aluminum block to calculate the weight of the water. Since the density of water is 1 g/cm³, the volume of water displaced by the aluminum block is equal to its volume, which is 172.22 cm³.

Finally, we can calculate the weight of the water by multiplying its volume by its density. The weight of the water is 172.22 cm³ * 1 g/cm³ = 172.22 g = 1.69 N (rounded to two decimal places).