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An acrobatic airplane performs a loop at an airshow. The centripetal acceleration the plane experiences is 14.7 m/s2. If it takes the pilot 45.0 seconds to complete the loop, what is the radius of the loop? Round your answer to the nearest whole number.
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\[\huge a_c=\frac{ 4\pi^2r }{ T^2 }\] \(a_c\) is the centripetal acceleration T is the time it takes to complete 1 revolution (AKA the period) r is the radius
U can also say a=omega^2*r
Omega=2*pi/T
some more flesh \(\huge a_c= \omega^2 \,r =\left( \frac{ 2\pi }{ T }\right)^2 r = \frac{ 4\pi^2 }{ T^2 }r\)
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