Answer:
\[{{m}_{1}}=14g\]
\[{{m}_{2}}=28g\]
\[{{F}_{1}}:{{F}_{2}}=?\]
\[F=ma\]
Slope of the velocity-time graph gives acceleration.
\[\Rightarrow \] slope \[(a)=\tan \theta \]
\[\Rightarrow F=m\times \tan \theta \]
\[\Rightarrow \frac{{{F}_{1}}}{{{F}_{2}}}=\frac{{{m}_{1}}\tan {{\theta }_{1}}}{{{m}_{2}}\tan {{\theta }_{2}}}\]
\[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{14\tan 45{}^\circ }{25\tan 30{}^\circ }\]
\[=\frac{\sqrt{3}}{2}\]
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