A jet weighs 3260kg and can withstand a max acceleration of "plus 7.33 g's". what is the net force acting on the jet?
would I put g as (-9.81) or just (9.81)
3 years ago
$F_{net}=m.a=3260\left(kg\right)\times7.33\times9.81\left(\frac{m}{s^2}\right)=234418N=234kN$Fnet=m.a=3260(kg)×7.33×9.81(ms2 )=234418N=234kN
3 years ago
Answered By Majid B
$F_{net}=m.a=3260\left(kg\right)\times7.33\times9.81\left(\frac{m}{s^2}\right)=234418N=234kN$Fnet=m.a=3260(kg)×7.33×9.81(ms2 )=234418N=234kN