Hi! Let me help you!
a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0
a = [(2t)/(4+t)]/t <---- working equation
a = {[2(5)]/9}/5 <---- cancel 5
a = 2/9 ft/s^2 <---- Answer
Right. You are true. The direction of the electric field is defined to be
the direction of the force on a small positive charge placed in the field.
Answer:
C. Angle of Attack.
Explanation:
The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.
And thus, we have
Lift = Weight
