Question

# Convert 1 dyne into newton.

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## Step 1 : Deriving CGS unit of force. $Force=mass\left(m\right)×acceleration\phantom{\rule{0ex}{0ex}}CGSunitofmass=grams\left(g\right)\phantom{\rule{0ex}{0ex}}CGSunitofacceleration=cm}{{s}^{2}}\phantom{\rule{0ex}{0ex}}CGSunitofForce=g×cm}{{s}^{2}}\phantom{\rule{0ex}{0ex}}=gcm}{{s}^{2}}\phantom{\rule{0ex}{0ex}}=dyne$Step 2 :Deriving SI unit of force $Force=mass\left(m\right)×accelaration\phantom{\rule{0ex}{0ex}}SIunitofmass=ki\mathrm{log}ram\left(kg\right)\phantom{\rule{0ex}{0ex}}SIunitofacceleration=m}{{s}^{2}}\phantom{\rule{0ex}{0ex}}SIunitofForce=kg×m}{{s}^{2}}\phantom{\rule{0ex}{0ex}}=kgm}{{s}^{2}}\phantom{\rule{0ex}{0ex}}=newton$Step 3 :Conversion of 1 dyne into newton $1dyne=1g×1cm}{{s}^{2}}\phantom{\rule{0ex}{0ex}}1gram={10}^{-3}kg\phantom{\rule{0ex}{0ex}}1cm={10}^{-2}m\phantom{\rule{0ex}{0ex}}1dyne={10}^{-3}kg×{10}^{-2}m}{{s}^{2}}\phantom{\rule{0ex}{0ex}}={10}^{-5}kgm}{{s}^{2}}\phantom{\rule{0ex}{0ex}}={10}^{-5}newton$Hence,$1dyne={10}^{-5}newton.$

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