Answer:
hello your question is incomplete attached below is the missing part
answer : short period oscillations frequency = 0.063 rad / sec
phugoid oscillations natural frequency ( 
) = 4.27 rad/sec
Explanation:
first we have to state the general form of the equation
= 
where :
comparing the general form with the given equation
= 18.2329
hence the short period oscillation frequency ( 
) = 0.063 rad/sec
phugoid oscillations natural frequency ( ) = 4.27 rad/sec
Answer:
W = 0.842 J
Explanation:
To solve this exercise we can use the relationship between work and kinetic energy
W = ΔK
In this case the kinetic energy at point A is zero since the system is stopped
W = K_f (1)
now let's use conservation of energy
starting point. Highest point A
Em₀ = U = m g h
Final point. Lowest point B
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
mg h = K
to find the height let's use trigonometry
at point A
cos 35 = x / L
x = L cos 35
so at the height is
h = L - L cos 35
h = L (1-cos 35)
we substitute
K = m g L (1 -cos 35)
we substitute in equation 1
W = m g L (1 -cos 35)
let's calculate
W = 0.500 9.8 0.950 (1 - cos 35)
W = 0.842 J
Answer:
A
Explanation:
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The correct option is A.
