In an automobile collision, a 44-kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.1
0-second time interval. What is the magnitude of the average force exerted on the passenger during this time?
1 answer:
Answer:
6,600N
Explanation:
According to second law of motion, Force = mass × acceleration
If acceleration = change in velocity/time = 15/0.10
Acceleration = 150m/s²
Given mass = 44kg
Force = 44× 150
Force = 6,600N
Magnitude of the average force exerted on the passenger during this time is 6,600N
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d= 250s
t= 17s
a= d/t
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= 4.7
Answer:
51kg
Explanation:
w=FxS
2500=Fx5
F=2500/5=500
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