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Find the terminal velocity of a spherical bacterium (diameter \(2.00 \mu \mathrm{m}\) ) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be \(1.10 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The terminal velocity of the spherical bacterium falling in water can be calculated using the following steps: 1. Calculate the volume of the bacterium: \(V = \frac{4}{3} \pi (1.00 \times 10^{-6} m)^3\) 2. Calculate the mass of the bacterium: \(m = (1.10 \times 10^{3} \mathrm{kg}/ \mathrm{m}^{3})\frac{4}{3} \pi (1.00 \times 10^{-6} m)^3\) 3. Calculate the weight of the bacterium: \(W = m(9.81 m/s^{2})\) 4. Find the terminal velocity: \(v = \frac{W}{6\pi (8.90 \times 10^{-4} Pa \cdot s)(1.00 \times 10^{-6} m)}\) Substitute the known values and solve for the terminal velocity \(v\).

Step by step solution

01

Calculate the volume of the bacterium

To calculate the volume of the spherical bacterium, we will use the formula for the volume of a sphere: \[V = \frac{4}{3} \pi r^3\] where \(r\) is the radius of the sphere. We have the diameter \(2.00 \mu m\) given, so the radius is half of that: \(1.00 \mu m\). Converting it to meters: \(r = 1.00 \times 10^{-6} m\). Now, we can plug this value into the formula to calculate the volume: \[V = \frac{4}{3} \pi (1.00 \times 10^{-6} m)^3\]
02

Calculate the mass of the bacterium

With the density of the bacterium \(\rho\) and volume \(V\) we found in step 1, we can calculate the mass of the bacterium using the following formula: \[m = \rho V\] Substitute the given values and the calculated volume to find the mass: \[m = (1.10 \times 10^{3} \mathrm{kg}/ \mathrm{m}^{3})\frac{4}{3} \pi (1.00 \times 10^{-6} m)^3\]
03

Calculate the weight of the bacterium

Now that we have calculated the mass of the bacterium, we can find its weight by multiplying its mass by the gravitational constant: \[W = mg\] Substitute the mass from step 2 and the gravitational constant \(g = 9.81m/s^{2}\) to find the weight: \[W = m(9.81 m/s^{2})\]
04

Find the terminal velocity

At terminal velocity, the drag force equals the weight of the bacterium. We will use the following equation to calculate the drag force \(F_d\): \[F_d = 6\pi \eta rv\] where \(\eta\) is the dynamic viscosity of the fluid, \(r\) is the radius of the bacterium, and \(v\) is the velocity. The terminal velocity will be reached when the drag force is equal to the weight of the bacterium (from Step 3). \[F_d = W\] \[6\pi \eta rv = W\] Now solve for the terminal velocity \(v\): \[v = \frac{W}{6\pi \eta r}\] Substitute the known values for the weight, viscosity, and radius of the bacterium: \[v = \frac{W}{6\pi (8.90 \times 10^{-4} Pa \cdot s)(1.00 \times 10^{-6} m)}\] Now we have the terminal velocity of the spherical bacterium falling in water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drag Force
In the context of a spherical bacterium falling through water, drag force plays a crucial role in determining its terminal velocity. The drag force is a type of frictional force that opposes the motion of an object through a fluid (like water) and is directly connected to the object's velocity. It arises due to differences in fluid pressure and the viscosity of the fluid around the different parts of the object. With regards to the bacterium, as it moves through water, it experiences resistance which increases with its speed, until the drag force becomes equal to the gravitational force pulling it downwards.

If you picture a bacterium descending through the water, the smoother and more spherical the bacterium, the more predictable the drag force becomes. In the case of the spherical bacterium mentioned in the exercise, Stokes’ law can be used to calculate the drag force exerted on it, which will be further explained in the relevant section below. Understanding drag force is vital because it contributes to determining the terminal velocity, which is the constant speed that the bacterium will reach and maintain as it falls through the water.
Bacterium Density
The density of a bacterium, or any object, is defined as its mass divided by its volume. It is a critical property that affects how the bacterium behaves when subjected to forces in a fluid. When we speak of bacterium density in the exercise, it implies a comparison between the mass of the bacterium and the volume of space it occupies, given in units of kilograms per cubic meter \(\mathrm{kg/m^3}\).

In this example, the density of the bacterium is specified as \(1.10 \times 10^3 \mathrm{kg/m^3}\). This value informs us that the bacterium is denser than water \(\mathrm{density~of~water} = 1000 \mathrm{kg/m^3}\), which means it will sink rather than float. The procedure for finding the bacterium's mass, and subsequently its weight, pivots on knowing its density and volume. This is pivotal because the weight of the bacterium is used to determine when the drag force has balanced it, resulting in terminal velocity.
Stokes' Law
Stokes' law is an equation that relates the drag force experienced by a spherical object moving through a viscous fluid to its velocity, radius, and the fluid's viscosity. It is represented by the formula \[F_d = 6\pi \eta rv\], where \(F_d\) is the drag force, \(r\) is the radius of the sphere, \(v\) is the velocity, and \(\eta\) is the dynamic viscosity of the fluid.

Importance of Stokes’ Law in Terminal Velocity

In the context of the problem, Stokes' law allows us to calculate the drag force that counterbalances the weight of the bacterial cell at terminal velocity. Terminal velocity occurs when the drag force equals the weight, indicating that no further acceleration takes place, and the falling object moves at a steady speed. By incorporating the known values, including the dynamic viscosity of water, the radius of the bacterium, and its weight, Stokes' law can be rearranged to solve for the terminal velocity of the bacterium.

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