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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.

User Whitehawk
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1 Answer

3 votes

Answer:

r=1,252.3528m

Step-by-step explanation:


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

User Kar
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