A 44.5 kg parachutist lands moving straight downward with a speed of 3.85 m/s.
(a) If the parachutist comes to rest with constant acceleration over a distance of 0.900 m, what force does the ground exerts on her?
N
(b) If the parachutist comes to rest over a longer distance, is the magnitude of the force exerted by the ground greater than, less than, or the same as in part (a)?
physics!!! please help!!!?
2016-10-28 9:23 pm
回答 (2)
2016-10-28 10:18 pm
a) Acceleration = (v^2/2d) = (3.85^2/1.8m) = -8.235m/sec^2.
Force from ground = (ma) = (44.5 x 8.235) = 366.457N.
b) Less force (longer distance to accelerate over).
Force from ground = (ma) = (44.5 x 8.235) = 366.457N.
b) Less force (longer distance to accelerate over).
2016-10-29 2:26 am
For impulse we have dP = m dV = Favg dT; so Favg = m dV/dT = m V/(s/V) = 44.5*3.85/(.9/3.85) = 733 N. ANS. a. dT = s/Vavg = .9/3.85 is the impact interval.
If s goes to S = 2s we have favg/Favg = (s/S) = 1/2 so favg < Favg and lengthening the impact distance will lower the average force. ANS b.
NOTE: This is the purpose of air bags, to lengthen the impact distance and reduce the force.
If s goes to S = 2s we have favg/Favg = (s/S) = 1/2 so favg < Favg and lengthening the impact distance will lower the average force. ANS b.
NOTE: This is the purpose of air bags, to lengthen the impact distance and reduce the force.
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