A 10cm times 10cm times 10cm block of steel (rho _steel= 7900kg/m^{3}) is suspended from a spring scale. The scale is in newtons.
Part A:What is the scale reading if the block is in air?
Part B:What is the scale reading after the block has been lowered into a beaker of oil (rho_oil=900kg/m^3) and is completely submerged?
Part A:
To find the scale reading when the block is in air, we need to calculate the weight of the block.
Step 1: Find the volume of the block.
The volume of the block is given by its dimensions, which are 10 cm x 10 cm x 10 cm.
Converting these lengths to meters, we have:
Volume = (10 cm) x (10 cm) x (10 cm) = (0.1 m) x (0.1 m) x (0.1 m) = 0.001 m³
Step 2: Calculate the weight of the block.
The weight is given by the equation:
Weight = Mass x Gravity
The mass of the block can be found using its volume and density, using the equation:
Mass = Density x Volume
Mass = (7900 kg/m³) x (0.001 m³) = 7.9 kg
Next, we need to determine the acceleration due to gravity, which is usually taken as 9.8 m/s².
Weight = 7.9 kg x 9.8 m/s² = 77.42 N
Therefore, the scale reading when the block is in air is 77.42 N.
Part B:
To find the scale reading when the block is completely submerged in oil, we need to calculate the buoyant force acting on the block.
Step 1: Find the volume of the block.
We already calculated the volume of the block in Part A, which is 0.001 m³.
Step 2: Calculate the weight of the displaced oil.
The weight of the displaced oil is given by the equation:
Weight of displaced oil = Density of oil x Volume of block x Gravity
Weight of displaced oil = (900 kg/m³) x (0.001 m³) x (9.8 m/s²) = 8.82 N
Step 3: Calculate the net force acting on the block.
The net force acting on the block can be found by subtracting the weight of the displaced oil from the weight of the block.
Net force = Weight of block - Weight of displaced oil
Net force = 77.42 N - 8.82 N = 68.60 N
Therefore, the scale reading after the block has been lowered into a beaker of oil and is completely submerged is 68.60 N.
To determine the scale reading in both scenarios, we need to consider the buoyant force acting on the block in each case.
Part A: Scale Reading in Air
In this case, the block is immersed in air, and the buoyant force is negligible since the density of air is significantly lower than the density of steel. Therefore, the scale reading would simply be equal to the weight of the block.
To find the weight of the block, we need to calculate its volume and multiply it by the density of steel.
The volume of the block can be determined by multiplying its dimensions:
Volume = (10 cm) x (10 cm) x (10 cm) = 1000 cm^3 = 0.001 m^3
Now, we can calculate the weight of the block:
Weight = Volume x Density = 0.001 m^3 x 7900 kg/m^3 = 7.9 kg
Thus, the scale reading in air would be 7.9 N.
Part B: Scale Reading in Oil
In this case, the block is completely submerged in oil, and there will be a buoyant force acting on it. The buoyant force is equal to the weight of the displaced oil, which is determined by the difference in densities between the block and the oil.
To calculate the buoyant force, we first need to find the volume of the block that is submerged in the oil. The submerged volume can be obtained by multiplying the dimensions of the block by the relevant fraction of its submerged height.
The submerged height can be determined by dividing the density of the steel block by the density of the oil:
Submerged height fraction = Density of steel / Density of oil = 7900 kg/m^3 / 900 kg/m^3 ≈ 8.78
Now, let's find the submerged volume:
Submerged volume = (10 cm) x (10 cm) x (10 cm) x Submerged height fraction
= 1000 cm^3 x 8.78 ≈ 8778 cm^3 = 0.008778 m^3
The buoyant force is equal to the weight of the displaced oil:
Buoyant force = Volume of displaced oil x Density of oil
= Submerged volume x Density of oil
= 0.008778 m^3 x 900 kg/m^3 = 7.90 N
Therefore, the scale reading after the block has been lowered into the beaker of oil and is completely submerged would be 7.90 N.
The volume of the block is .1^3 meters^3
weightinAir=density*volume*g
Now, in oil, buoyancy subtracts..
weight=weight in air-weight of oil displaced.
figure the weight of oil displaced the same way as above