# What will be the dimensional formula of momentum if force,acceleration&time are taken as fundamental

$F=\frac{dp}{dt}\phantom{\rule{0ex}{0ex}}dp=Fdt\phantom{\rule{0ex}{0ex}}thedimensionsymbolforforce=FfortimeTforaccnA\phantom{\rule{0ex}{0ex}}fromaboverelation\phantom{\rule{0ex}{0ex}}momentum=\left[FT{A}^{0}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}regards$

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