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(A) The formula for the angle of banking for a horizontal circular turn is given by $	an 	heta = \frac{v^2}{rg}$.Given: speed $v = 150 \, m/s$,banki

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$10.6$

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(A) The formula for the angle of banking for a horizontal circular turn is given by $	an 	heta = \frac{v^2}{rg}$.Given: speed $v = 150 \, m/s$,banking angle $	heta = 12^\circ$,and acceleration due to gravity $g = 10 \, m/s^2$.Rearranging the formula to solve for the radius $r$: $r = \frac{v^2}{g 	an 	heta}$.Substituting the values: $r = \frac{(150)^2}{10 	imes 	an 12^\circ}$.Using $	an 12^\circ \approx 0.2125$,we get $r = \frac{22500}{10 	imes 0.2125} = \frac{22500}{2.125} \approx 10588 \, m$.Converting to kilometers: $r \approx 10.6 \, km$.

An aircraft executes a horizontal loop with a speed of $150 \, m/s$ with its wings banked at an angle of $12^\circ$. The radius of the loop is .......... $km$. $(g = 10 \, m/s^2)$

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