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If the jet is moving at a speed of 1040 km/h at the lowest point of the loop, determine the minimum radius of the circle so that the centripetal acceleration at the lowest point does not exceed 6.8 g's.

Respuesta :

noahandersen2004

Answer:

r=1,252.3528m

Explanation:

[tex]F_c=ma_c=\frac{mv^2}{r}\\a_c=\frac{v^2}{r}\\r=\frac{v^2}{a_c}\\r=\frac{((1040) (1000)m/(60)(60)s)^2}{(6.8g)m/s^2}\\r=\frac{((1040000)m/(3600)s)^2}{(6.8 \cdot 9.8)m/s^2}\\r=\frac{(288.889m/s)^2}{66.64m/s^2}\\r=\frac{83,456.790m^2/s^2}{66.64m/s^2}\\r=\frac{83,456.790m}{66.64}\\r=1,252.3528m[/tex]

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