In this particular situation, the resistive force, $$ \dot{F}_{R} $$, the air exerts on the falling raindrop is given by $$ \dot{F}_{R}=-b \dot{v} $$, where $$ \dot{v} $$ is the velocity of the raindrop. b is a constant that depends on the size and shape of the falling object but not its material. f) Determine the numerical value of the constant b in the expression for the air resistance force on the raindrop of this particular situation. $$ b= $$ g) What if the raindrop is replaced by a plastic pellet of exactly the same shape and size, but with a density, 2.10 times the density of water, what would be the pellet's weight? $$ \dot{F}_{g}= $$ g) What would be the pellet's terminal velocity? $$ \dot{v}_{t}= $$ Question Help: D Post to forum

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To solve the problem, we need to break it down into parts based on the questions asked.

### Part f: Determine the numerical value of the constant $$ b $$

The resistive force $$ \dot{F}_{R} $$ is given by:

$$
\dot{F}_{R} = -b \dot{v}
$$

To find the value of $$ b $$, we need additional information about the raindrop, such as its size, shape, and the conditions under which the resistive force is measured. If we assume that we have the terminal velocity $$ \dot{v}_{t} $$ and the weight of the raindrop $$ \dot{F}_{g} $$, we can set up the equation at terminal velocity where the resistive force equals the weight of the raindrop:

$$
\dot{F}_{g} = b \dot{v}_{t}
$$

If we have values for $$ \dot{F}_{g} $$ and $$ \dot{v}_{t} $$, we can rearrange this to find $$ b $$:

$$
b = \frac{\dot{F}_{g}}{\dot{v}_{t}}
$$

### Part g: Weight of the plastic pellet

The weight of an object is given by the formula:

$$
\dot{F}_{g} = \rho V g
$$

Where:
- $$ \rho $$ is the density of the object,
- $$ V $$ is the volume of the object,
- $$ g $$ is the acceleration due to gravity (approximately $$ 9.81 \, \text{m/s}^2 $$).

Given that the density of the plastic pellet is $$ 2.10 $$ times the density of water, we can express the weight of the pellet as:

$$
\dot{F}_{g} = (2.10 \rho_{water}) V g
$$

### Part h: Terminal velocity of the plastic pellet

The terminal velocity can be found using the same principle as before. At terminal velocity, the weight of the pellet equals the resistive force:

$$
\dot{F}_{g} = b \dot{v}_{t}
$$

Substituting the expression for $$ \dot{F}_{g} $$:

$$
(2.10 \rho_{water}) V g = b \dot{v}_{t}
$$

From this, we can solve for the terminal velocity $$ \dot{v}_{t} $$:

$$
\dot{v}_{t} = \frac{(2.10 \rho_{water}) V g}{b}
$$

### Summary

To proceed with the calculations, we need the following:
1. The density of water $$ \rho_{water} $$ (approximately $$ 1000 \, \text{kg/m}^3 $$).
2. The volume $$ V $$ of the raindrop or pellet.
3. The terminal velocity $$ \dot{v}_{t} $$ of the raindrop to find $$ b $$.

If you provide the necessary values, I can perform the calculations for $$ b $$, $$ \dot{F}_{g} $$, and $$ \dot{v}_{t} $$.
To determine the constant $$ b $$, you need the raindrop's terminal velocity and its weight. Once you have those, use the formula $$ b = \frac{\dot{F}_{g}}{\dot{v}_{t}} $$. For the plastic pellet, its weight is $$ \dot{F}_{g} = 2.10 \times \rho_{water} \times V \times g $$. The pellet's terminal velocity is $$ \dot{v}_{t} = \frac{2.10 \times \rho_{water} \times V \times g}{b} $$.

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