Question #751d5

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Answer 1 · 2017-03-17T10:56:55

The ratio is $= 0.54$ $=0.54$

Let

mass of plane $= m k g$ $=mkg$

speed of plane $= v m s^{- 1}$ $=vms^-1$

radius of circle $= r m$ $=rm$

The centripetal force is $F = \frac{m v^{2}}{r} N$ $F=(mv^2)/r N$

The weight of the plane $W = m g N$ $W=mg N$

The ratio

$\frac{F}{W} = \frac{m v^{2}}{r} \cdot \frac{1}{m g}$ $F/W=(mv^2)/r*1/(mg)$

$= \frac{v^{2}}{r g} = \frac{650^{2}}{80000 \cdot 9.8}$ $=v^2/(rg)=650^2/(80000*9.8)$

$= 0.54$ $=0.54$