Answer:
1626.4 N
Explanation:
Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?
The parameters to be considered are:
Distance S = 3m
Time t = 0.55s
Since the man started from rest, initial velocity u = 0
Using second equation of motion
S = Ut + 1/2at^2
3 = 1/2 × a × 0.55^2
3 = 1/2 × a × 0.3025
a = 3/ 0.15125
a = 19.83 m/s^2
Force = mass × acceleration
Force = 82 × 19.83
Force = 1626.4 N
Therefore, the force that water exerted on him is 1626.4 N
Answer:
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Explanation:
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Answer:
2.85 rad/s
Explanation:
5 cm = 0.05 m
20 g = 0.02 kg
When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:
So the total moment of inertia of the system of 2 objects after the drop is:
From here we can apply the law of angular momentum conservation to calculate the post angular speed
Answer: The answer is A
Explanation: The bumper is the first part of an automobile to be impacted when in a head-on accident
