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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, $q = \frac{\gamma+1}{2}$.
A projectile is fired at $45°$ to the horizontal. The elevation angle of the projectile at its highest point, as seen from the point of projection, is:
At max height $H = \dfrac{u^2}{4g}$; horizontal distance $= \dfrac{R}{2} = \dfrac{u^2}{2g}$
Elevation angle: $\tan\phi = \dfrac{H}{R/2} = \dfrac{u^2/(4g)}{u^2/(2g)} = \dfrac{1}{2}$
$\phi = \mathbf{\tan^{-1}\!\left(\dfrac{1}{2}\right)}$
A stone is dropped from rest from the top of a $20\text{ m}$ tower. Its speed on hitting the ground is ($g = 10\text{ m/s}^2$):
Using $v^2 = u^2 + 2gh = 0 + 2(10)(20) = 400$
$v = \sqrt{400} = \mathbf{20\text{ m/s}}$
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