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The position of a 2.75×105n training helicopter under test is given by r⃗ =(0.020m/s3)t3i^+(2.2m/s)tj^−(0.060m/s2)t2k^. part a find the net force on the helicopter at t=5.0s.

Justin Parker

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Daniel King on August 17, 2018

F (t) = 1.7⋅10^4i^-3.4⋅10^3kHere we have the components in the x-direction and the y-axis and directionFx = 17000 NFz = 3400 Nla resultant force is given byF = sqrt(Fx2 + Fz2)= sqrt(170002 + 34002 )= 17336.67 NNet force on the helicopter at t = 5.0 s is 17336.67 N


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