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:
The natural frequency = 50 rad/s = 7.96 Hz
Damping ratio = 0.5
Explanation:
The natural frequency is calculated in this manner
w = √(k/m)
k = spring constant = 5 N/m
m = mass = 2 g = 0.002 kg
w = √(5/0.002) = 50 rad/s
w = 2πf
50 = 2πf
f = 50/(2π) = 7.96 Hz
Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5
Answer:
The gauge pressure is 
Explanation:
From the question we are told that
The height of the water contained is 
The height of liquid in the cylinder is 
At the bottom of the cylinder the gauge pressure is mathematically represented as

Where
is the pressure of water which is mathematically represented as

Now
is the density of water with a constant values of 
substituting values


While
is the pressure of oil which is mathematically represented as

Where
is the density of oil with a constant value

substituting values


Therefore

