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Question-12331




Question Number 12331 by tawa last updated on 19/Apr/17
Answered by mrW1 last updated on 19/Apr/17
m(v^2 /r)≤μmg  μ≥(v^2 /(rg))=(5^2 /(8×9.81))=0.32
$${m}\frac{{v}^{\mathrm{2}} }{{r}}\leqslant\mu{mg} \\ $$$$\mu\geqslant\frac{{v}^{\mathrm{2}} }{{rg}}=\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{8}×\mathrm{9}.\mathrm{81}}=\mathrm{0}.\mathrm{32} \\ $$
Commented by tawa last updated on 19/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by By, Dismas RB last updated on 20/Apr/17
data   verocity=5.0m/s  distance=8.0m  required
$${data} \\ $$$$\:{verocity}=\mathrm{5}.\mathrm{0}{m}/{s} \\ $$$${distance}=\mathrm{8}.\mathrm{0}{m} \\ $$$${required} \\ $$

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