A paperweight, when weighed in air, has a weight of W = 6.95 N. When completely immersed in water, however, it has a weight of Win water = 4.26 N. Find the volume of the paperweight.
Would I multiply the two values of weight and then by the density of water?
4.26=weightinair-forcebouyancy
4.26=6.95-desnisitywater*g*volume
solve for volume
So would I set it up as
4.26= 6.95 - 1000(9.8)V ?
weight of water displaced = 6.95 - 4.26 = 2.69 N
2.69 N = Vol * 1000Kg/m^3 * 9.81 m/s^2
so
Yes
To find the volume of the paperweight, you can use the principle of buoyancy. When an object is immersed in a fluid, it experiences an upward force called buoyant force, which is equal to the weight of the fluid displaced by the object.
So, let's break down the problem step by step:
Step 1: Find the difference in weight when the paperweight is submerged in water.
The difference in weight between the paperweight in air and in water represents the buoyant force. So, you can calculate it using the equation:
Win water = W - Force of buoyancy
Step 2: Calculate the force of buoyancy.
The force of buoyancy is equal to the weight of the fluid displaced by the paperweight. According to Archimedes' principle, the buoyant force is also equal to the weight of the fluid displaced. We can express this as:
Force of buoyancy = density of water * gravitational acceleration * Volume
Step 3: Substitute the given values into the equation.
You can substitute the values into the equation and solve for the volume:
Win water = W - (density of water * gravitational acceleration * Volume)
4.26 N = 6.95 N - (1000 kg/m^3 * 9.8 m/s^2 * Volume)
Step 4: Solve for Volume.
Rearrange the equation to solve for Volume:
Volume = (W - Win water) / (density of water * gravitational acceleration)
Volume = (6.95 N - 4.26 N) / (1000 kg/m^3 * 9.8 m/s^2)
Volume = 0.00276 m^3
Therefore, the volume of the paperweight is 0.00276 cubic meters.