Linearize the following equations into the form y = mx + c: T = sqrt((m + m0/3)/(kg)), v = l/(4f) - c, P = P0e^(-kh)

Let $y = T^2$ and $x = m$.
The linearized equation is:

$$\boxed{T^2 = \left(\frac{1}{kg}\right)m + \left(\frac{m_0}{3kg}\right)}$$

Here, the slope $m_{slope} = \frac{1}{kg}$ and the y-intercept $c_{intercept} = \frac{m_0}{3kg}$.

2. For the equation $v = \frac{l}{4f} - c$

$$l = 4f(v+c)$$ $$l = 4fv + 4fc$$

We want to find $v$ and $c$. If we plot $l$ against $\frac{1}{f}$, we can achieve a linear relationship.

$$l = \frac{v}{4} \left(\frac{1}{f}\right) - c$$

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

$$\boxed{l = \left(\frac{v}{4}\right)\left(\frac{1}{f}\right) - c}$$

Here, the slope $m_{slope} = \frac{v}{4}$ and the y-intercept $c_{intercept} = -c$.

3. For the equation $P = P_0 e^{-kh}$

$$\ln(P) = \ln(P_0) + \ln(e^{-kh})$$ $$\ln(P) = \ln(P_0) - kh$$

Let $y = \ln(P)$ and $x = h$.
The linearized equation is:

$$\boxed{\ln(P) = -kh + \ln(P_0)}$$

Here, the slope $m_{slope} = -k$ and the y-intercept $c_{intercept} = \ln(P_0)$.

4. For the equation $\lambda^2 (\eta - \beta) = k$

$$\eta = \frac{k}{\lambda^2} + \frac{\lambda^2 \beta}{\lambda^2}$$ $$\eta = k \left(\frac{1}{\lambda^2}\right) + \beta$$

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

$$\boxed{\eta = k \left(\frac{1}{\lambda^2}\right) + \beta}$$

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