An object has a weight of 0.852 N. It is suspended from a scale, which reads 0.749 N when the piece is submerged in water. What is the volume of the piece of metal?

Buoyancy force = 0.103 N

0.103 N = V*(water density)*g
0.852 N = V*(object density)*g

(Object density)/(water density) = 8.27
Object density = 8.27*10^3 kg/m^3
Object volume = V = 0.852/[9.81*8.27*10^3)]
= 1.05*10^5 m^3 = 10.5 cm^3

To find the volume of the piece of metal, you can use Archimedes' principle, which states that the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.

Let's break down the problem and calculate step by step:

Step 1: Determine the weight of the object in air.
The weight of the object in air is given as 0.852 N.

Step 2: Determine the weight of the object when submerged in water.
The scale reads 0.749 N when the object is submerged in water.

Step 3: Calculate the buoyant force.
By Archimedes' principle, the buoyant force acting on the object is equal to the weight of the water displaced by the object. So, the buoyant force is 0.749 N.

Step 4: Calculate the weight of the water displaced.
The weight of the water displaced is equal to the weight of the object in air minus the weight of the object in water:
0.852 N - 0.749 N = 0.103 N.

Step 5: Calculate the volume of water displaced.
Knowing that the density of water is 1000 kg/m³ and the weight of the water displaced is 0.103 N, we can use the formula:
Weight = density * volume * gravity.
Substituting the values:
0.103 N = 1000 kg/m³ * volume * 9.8 m/s².

Rearranging the formula to solve for volume:
volume = weight / (density * gravity) = 0.103 N / (1000 kg/m³ * 9.8 m/s²) = 0.0106 m³.

Therefore, the volume of the piece of metal is approximately 0.0106 cubic meters.