Linearize the following equations to the form y=mx+c and identify the slope and y-intercept: a) F = Q2/(4πε₀r²); b) R = R₀(1+αθ); c) h−H = ut + ½gt² with u=0.

To determine $\epsilon_0$, we can plot $F$ against $\frac{1}{r^2}$.

$$F = \left(\frac{Q^2}{4\pi \epsilon_0}\right) \left(\frac{1}{r^2}\right)$$

Let $y = F$ and $x = \frac{1}{r^2}$.
The linearized equation is:

$$\boxed{F = \left(\frac{Q^2}{4\pi \epsilon_0}\right)\left(\frac{1}{r^2}\right)}$$

Here, the slope $m_{slope} = \frac{Q^2}{4\pi \epsilon_0}$ and the y-intercept $c_{intercept} = 0$.

2. For the equation $R = R_0(1 + \alpha \theta)$ ()

To obtain $R_0$ and $\alpha$, we can plot $R$ against $\theta$.

$$R = (R_0 \alpha)\theta + R_0$$

Let $y = R$ and $x = \theta$.
The linearized equation is:

$$\boxed{R = (R_0 \alpha)\theta + R_0}$$

Here, the slope $m_{slope} = R_0 \alpha$ and the y-intercept $c_{intercept} = R_0$.

3. For the equation $h - H = ut + \frac{1}{2}gt^2$ with $u=0$ ()

$$h - H = (0)t + \frac{1}{2}gt^2$$ $$h - H = \frac{1}{2}gt^2$$

To obtain $H$ and $g$, we can plot $h$ against $t^2$.

$$h = \left(\frac{1}{2}g\right)t^2 + H$$

Let $y = h$ and $x = t^2$.
The linearized equation is:

$$\boxed{h = \left(\frac{1}{2}g\right)t^2 + H}$$

Here, the slope $m_{slope} = \frac{1}{2}g$ and the y-intercept $c_{intercept} = H$.

4. For the equation $PV = k$ ()

To determine $k$, we can plot $P$ against $\frac{1}{V}$.

$$P = k \left(\frac{1}{V}\right)$$

Let $y = P$ and $x = \frac{1}{V}$.
The linearized equation is:

$$\boxed{P = k \left(\frac{1}{V}\right)}$$

Here, the slope $m_{slope} = k$ and the y-intercept $c_{intercept} = 0$.