Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height
of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
1 answer:
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Answer:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s
from equation of motion we know that
where d_1 is distance covered in time t1
so=
where d_2 is distance covered in time t2
distance cover is = 213.31 - 110.78 = 102.53 m
Answer:2.5 m/s
37.5KJ
Explanation:
Let be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.
Conserving momentum
Therefore Final velocity of system is 2.5m/s
(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy