- #1
kenau_reveas
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When the parachute is fully open, the effective drag coefficient of the sky diver plus parachute increases to 60.0 kg/m. What is the drag force F_drag acting on the sky diver immediately after she has opened the parachute?
i know that F_drag = Dv^2
here D is given which is 60 kg/m. and i found v by the equation v= square root of ((m.g)/D)
so i got 13.06 for V^2
and then i put the values in the equation F_drag = D.V^2
= 783. 6
is this right?
i know that F_drag = Dv^2
here D is given which is 60 kg/m. and i found v by the equation v= square root of ((m.g)/D)
so i got 13.06 for V^2
and then i put the values in the equation F_drag = D.V^2
= 783. 6
is this right?