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Wavelength associated with freely falling body:
As it falls, velocity increases ⇒ de Broglie wavelength λ = h/mv decreases.
The exponent must be dimensionless. Thus, $[\alpha] = \frac{[x^2]}{[kT]}$.
Since $[kT]$ has dimensions of energy $[ML^2T^{-2}]$, $[\alpha] = \frac{[L^2]}{[ML^2T^{-2}]} = [M^{-1}T^2]$.
Force $F$ has the same dimensions as $\alpha\beta$.
$[\beta] = \frac{[F]}{[\alpha]} = \frac{[MLT^{-2}]}{[M^{-1}T^2]} = [M^2 L T^{-4}]$.
The horizontal range and maximum height of a projectile are equal. The angle of projection is:
$R = H \Rightarrow \dfrac{u^2\sin2\theta}{g} = \dfrac{u^2\sin^2\theta}{2g}$
$4\sin\theta\cos\theta = \sin^2\theta \Rightarrow 4\cos\theta = \sin\theta \Rightarrow \tan\theta = 4$
$\theta = \mathbf{\tan^{-1}(4)}$
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