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Calculate how many g's (gravity accelerations) the glider pilot when turning the horizontal circles of radius 139 m flying at 126 km/h. Centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius of rotation.

Consider a vertically acting gravity acceleration g=9.81 ms^-1.

Final Answer:

x =  1.3443 g

Step-by-step explanation:

$$\begin{gathered} r=139\text{m} \\ v=126\text{km/h}\to \text{m/s}=126:3.6\text{m/s}=35\text{m/s} \\ g=9.81{\text{m/s}}^2 \\ {a}_1=\frac{v^2}{r}=\frac{35^2}{139}=\frac{1225}{139}\doteq 8.8129{\text{m/s}}^2 \\ {a}^2=g^2+a_C^2 \\ x=\frac{a}{g} \\ x=\frac{\sqrt{g^2+a_1^2}}{g}=\frac{\sqrt{9.81^2+8.8129^2}}{9.81}=1.3443\text{g} \end{gathered}$$

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