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If the force is given by $F=a t+b t^2$ with $t$ as time. The dimensions of $a$ and $b$ are
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Verified Answer
The correct answer is:
$\left[\mathrm{MLT}^{-3}\right],\left[\mathrm{MLT}^{-4}\right]$
Dimension of $a t=$ Dimension of $F$
$[a t]=[F]$
$[a]=\left[\frac{F}{t}\right]$
$[a]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~T}}\right]$
$[a]=\left[\mathrm{MLT}^{-3}\right]$
Dimension of $b t^2=$ Dimension of $F$
$\left[b t^2\right]=[F]$
$[b]=\left[\frac{F}{t^2}\right]$
$[b]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~T}^2}\right]$
$[b]=\left[\mathrm{MLT}^{-4}\right]$
$[a t]=[F]$
$[a]=\left[\frac{F}{t}\right]$
$[a]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~T}}\right]$
$[a]=\left[\mathrm{MLT}^{-3}\right]$
Dimension of $b t^2=$ Dimension of $F$
$\left[b t^2\right]=[F]$
$[b]=\left[\frac{F}{t^2}\right]$
$[b]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~T}^2}\right]$
$[b]=\left[\mathrm{MLT}^{-4}\right]$
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