Determine the slope of the graph

2(b) and list the corresponding time ($t_i$) values from the table in Fig. 2(a).

From the previous steps, we have:

• For $d_1 = 1.0 \text{ cm}$, $V_1 \approx 0.5 \text{ cm}^3$. The corresponding time is $t_1 = 10 \text{ s}$.
• For $d_2 = 2.0 \text{ cm}$, $V_2 \approx 1.5 \text{ cm}^3$. The corresponding time is $t_2 = 20 \text{ s}$.
• For $d_3 = 3.0 \text{ cm}$, $V_3 \approx 3.0 \text{ cm}^3$. The corresponding time is $t_3 = 30 \text{ s}$.
• For $d_4 = 4.0 \text{ cm}$, $V_4 \approx 5.0 \text{ cm}^3$. The corresponding time is $t_4 = 40 \text{ s}$.
• For $d_5 = 5.0 \text{ cm}$, $V_5 \approx 7.5 \text{ cm}^3$. The corresponding time is $t_5 = 50 \text{ s}$.

The data points for the graph with $V$ on the vertical axis and $t$ on the horizontal axis are:

$$(10 \text{ s}, 0.5 \text{ cm}^3), (20 \text{ s}, 1.5 \text{ cm}^3), (30 \text{ s}, 3.0 \text{ cm}^3), (40 \text{ s}, 5.0 \text{ cm}^3), (50 \text{ s}, 7.5 \text{ cm}^3)$$

The slope of a $V$ vs $t$ graph represents the rate of change of volume with respect to time. Since the relationship between $V$ and $t$ is not perfectly linear, we will calculate the average slope using the first and last data points.
The formula for the slope ($m$) is:

$$m = \frac{\Delta V}{\Delta t} = \frac{V_5 - V_1}{t_5 - t_1}$$

Substitute the values:

$$m = \frac{7.5 \text{ cm}^3 - 0.5 \text{ cm}^3}{50 \text{ s} - 10 \text{ s}}$$ $$m = \frac{7.0 \text{ cm}^3}{40 \text{ s}}$$ $$m = 0.175 \text{ cm}^3/\text{s}$$

The slope of the graph is $\text{0.175 cm}^3/\text{s}$.