A thin spherical shell of mass 0.400 kg and diameter 0.190 m is filled with alcohol (ρ = 806 kg/m3). It is then released from rest on the bottom of a pool of water. Find the acceleration of the alcohol filled shell as it rises toward the surface of the water.
To find the acceleration of the alcohol-filled shell as it rises toward the surface of the water, we can use the principles of buoyancy and Newton's second law of motion.
Step 1: Calculate the buoyant force acting on the shell.
The buoyant force is given by the formula: Buoyant force = weight of the displaced fluid
The weight of the displaced alcohol is given by the formula: weight = mass * acceleration due to gravity
The volume of the shell can be calculated using its diameter: volume = 4/3 * π * (diameter/2)^3
Step 2: Calculate the weight of the alcohol-filled shell.
The weight of the alcohol-filled shell is given by the formula: weight = mass * acceleration due to gravity
Step 3: Calculate the net force acting on the alcohol-filled shell.
The net force is given by the formula: net force = weight of the alcohol-filled shell - buoyant force
Step 4: Calculate the acceleration of the alcohol-filled shell.
The acceleration is given by the formula: acceleration = net force / mass of the alcohol-filled shell
Now, let's calculate the acceleration step by step.
Step 1:
Volume = 4/3 * π * (0.190/2)^3
= 0.0611 m^3
Buoyant force = weight of the displaced fluid
= density of alcohol * volume * acceleration due to gravity
= 806 kg/m^3 * 0.0611 m^3 * 9.8 m/s^2
= 473 N
Step 2:
Weight of the alcohol-filled shell = mass of the shell * acceleration due to gravity
= 0.400 kg * 9.8 m/s^2
= 3.92 N
Step 3:
Net force = weight of the alcohol-filled shell - buoyant force
= 3.92 N - 473 N
= -469.08 N (upward direction)
Step 4:
Acceleration = net force / mass of the alcohol-filled shell
= -469.08 N / 0.400 kg
= -1173 m/s^2
Therefore, the acceleration of the alcohol-filled shell as it rises toward the surface of the water is -1173 m/s^2 (upward direction).
To find the acceleration of the alcohol-filled shell as it rises toward the surface of the water, we can use the concept of buoyancy.
Buoyancy is the upward force exerted on an object submerged in a fluid. It is equal to the weight of the fluid displaced by the object.
First, let's find the volume of the alcohol-filled shell. The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius.
Given that the diameter of the shell is 0.190 m, the radius is half of the diameter, so r = 0.190/2 = 0.095 m.
Now we can calculate the volume of the shell:
V = (4/3)π(0.095)^3 = 0.00126 m^3
Next, we need to find the weight of the alcohol-filled shell. The weight can be calculated using the equation W = mg, where m is the mass and g is the acceleration due to gravity.
The mass of the shell is given as 0.400 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
So, the weight of the shell is:
W = 0.400 kg × 9.8 m/s^2 = 3.92 N
Now, we need to find the weight of the displaced water. The weight of the displaced water is equal to the buoyant force acting on the shell.
The formula for the buoyant force is F_buoyant = ρ_fluid × V × g, where ρ_fluid is the density of the fluid (water in this case), V is the volume of the displaced fluid, and g is the acceleration due to gravity.
The density of alcohol is given as ρ_alcohol = 806 kg/m^3, so the density of the fluid is the same. The volume of the displaced fluid is equal to the volume of the shell.
Therefore, the weight of the displaced water is:
W_displaced = ρ_fluid × V × g = 806 kg/m^3 × 0.00126 m^3 × 9.8 m/s^2 = 9.9 N
Finally, the net force acting on the alcohol-filled shell as it rises is the difference between its weight and the weight of the displaced water:
Net force = W - W_displaced = 3.92 N - 9.9 N = -5.98 N
Since the net force is negative, it means that the downward force (shell's weight) is greater than the upward force (buoyant force). This results in an acceleration opposite to the direction of the net force.
Therefore, the acceleration of the alcohol-filled shell as it rises toward the surface of the water is 5.98 N, directed downward.