Net force (taking up as positive): $F_{net} = -mg - 0.2v^2 = -20 - 0.2v^2$

Using $ma = v\dfrac{dv}{dx}$: $2v\dfrac{dv}{dx} = -20 - 0.2v^2$

Separating variables: $\dfrac{v\,dv}{20+0.2v^2} = -\dfrac{dx}{2}$

Integrating from $v = 10$ to $v = 0$: let $k = 20 + 0.2v^2$, $dk = 0.4v\,dv$

$\dfrac{1}{0.4}\big[\ln(20) - \ln(40)\big] = -\dfrac{H}{2} \Rightarrow H = \dfrac{2\ln 2}{0.4} = \mathbf{5\ln 2}$

Average time between collisions $\tau$ is given by $\tau = \frac{\lambda}{v_{rms}}$, where $\lambda$ is the mean free path.

1. $\lambda \propto \frac{1}{n} \propto V$ (where $n$ is number density).
2. $v_{rms} \propto \sqrt{T}$.
3. For an adiabatic process, $TV^{\gamma-1} = \text{constant}$, so $T \propto V^{1-\gamma}$ and $\sqrt{T} \propto V^{\frac{1-\gamma}{2}}$.

Therefore, $\tau \propto \frac{V}{V^{\frac{1-\gamma}{2}}} = V^{1 - \frac{1-\gamma}{2}} = V^{\frac{2-1+\gamma}{2}} = V^{\frac{\gamma+1}{2}}$.
Hence, $q = \frac{\gamma+1}{2}$.

Option B is incorrect because 'give' should be the past participle 'given' in a passive voice construction.

• If \(p \leq \alpha\): Reject the null hypothesis \((H_0)\)
• If \(p > \alpha\): Fail to reject the null hypothesis

Roots: $x=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm i$. So $\alpha=-1+i, \beta=-1-i$.

In polar form: $|\alpha|=\sqrt{2}$, $\arg(\alpha)=\frac{3\pi}{4}$. So $\alpha=\sqrt{2}\,e^{i3\pi/4}$.

$\alpha^{15}=(\sqrt{2})^{15}e^{i\cdot45\pi/4}=2^{15/2}e^{i\pi/4}$ (since $45\pi/4=11\pi+\pi/4$, so $e^{i45\pi/4}=e^{i\pi/4}\cdot(-1)^{11}=-e^{i\pi/4}$... careful: $45/4=11.25$, $11\pi+\pi/4$, $e^{i(11\pi+\pi/4)}=e^{i\pi}\cdot e^{i\pi/4} \cdot (-1)^{10}... $)

More cleanly: $\alpha^{15}+\beta^{15}=2\,\text{Re}(\alpha^{15})=2(\sqrt{2})^{15}\cos\!\left(\frac{45\pi}{4}\right)$. Since $\cos(45\pi/4)=\cos(\pi/4+11\pi)=-\cos(\pi/4)=-\frac{1}{\sqrt{2}}$: $=2\cdot2^{15/2}\cdot(-\frac{1}{\sqrt{2}})=-2^{15/2+1-1/2}=-2^8=-\mathbf{256}$