Asked by ami
The position of a 2.75 *10^5 N training helicopter under test is given by
r = (0.020 m/s^3)(t^3)i + (2.2 m/s)(t)j - (0.060 m/s^2)(t^2) k
Find the net force on the helicopter at t =5.0s.
Express your answer in terms of i,j,k. Express your coefficient using two significant figures.
r = (0.020 m/s^3)(t^3)i + (2.2 m/s)(t)j - (0.060 m/s^2)(t^2) k
Find the net force on the helicopter at t =5.0s.
Express your answer in terms of i,j,k. Express your coefficient using two significant figures.
Answers
Answered by
TutorCat
just plug in 5 for t:
r = (0.020 m/s^3)(5^3)i + (2.2 m/s)(5)j - (0.060 m/s^2)(5^2) k
r = (0.020 m/s^3)(5^3)i + (2.2 m/s)(5)j - (0.060 m/s^2)(5^2) k
Answered by
TutorCat
1. take the second derivative of r to find acceleration.
2. then divide 2.75 *10^5 N by 9.8m/s^2 to find how much the helicopter weighs in kg
3. multiply your result from 1 with your result from 2
2. then divide 2.75 *10^5 N by 9.8m/s^2 to find how much the helicopter weighs in kg
3. multiply your result from 1 with your result from 2
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