A thin spherical shell of mass 0.450 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 Newton's second law of motion: F = ma, where F is the net force acting on the object, m is the mass of the object, and a is the acceleration.
First, let's find the net force acting on the alcohol-filled shell.
When the shell is submerged in water, there are two forces acting on it: buoyant force and the weight of the shell.
The buoyant force is given by Archimedes' principle: Fb = ρVg, where ρ is the density of the water, V is the volume of the shell, and g is the acceleration due to gravity.
The weight of the shell is given by: W = mg, where m is the mass of the shell, and g is the acceleration due to gravity.
Since the shell is spherical, its volume can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3, where r is the radius of the shell.
The radius of the shell is half the diameter, so r = 0.190 m / 2 = 0.095 m.
Now let's calculate the volume of the shell:
V = (4/3)π(0.095 m)^3 = 0.000257 m^3.
Substituting the values into the equation for the buoyant force:
Fb = ρVg = (806 kg/m^3)(0.000257 m^3)(9.8 m/s^2) = 2.02 N.
The weight of the shell:
W = mg = (0.450 kg)(9.8 m/s^2) = 4.41 N.
Now, let's calculate the net force:
F = Fb - W = 2.02 N - 4.41 N = -2.39 N.
The negative sign indicates that the net force is in the opposite direction of motion, which means the shell is being slowed down.
Finally, we can use Newton's second law to find the acceleration of the shell:
F = ma. Rearranging the equation, we get:
a = F/m.
Substituting the values:
a = (-2.39 N)/(0.450 kg) = -5.31 m/s^2.
Therefore, the acceleration of the alcohol-filled shell as it rises toward the surface of the water is -5.31 m/s^2.