Aquatic ecosystems consist of two different layers with respect to light. The photic zone is where light penetrates the water, a
2 answers:
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Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
Answer:
a) the maximum shear stress τ the bar is 16T
/πd³
b) the angle of twist between the ends of the bar is 16tL² / πGd⁴
Explanation:
Given the data in the question, as illustrated in the image below;
d is the diameter of the prismatic bar of length AB
t is the intensity of distributed torque
(a) Determine the maximum shear stress tmax in the bar
Maximum Applied torque T_max = tL
we know that;
shear stress τ = 16T/πd³
where d is the diameter
so
τ = 16T
/πd³
Therefore, the maximum shear stress τ the bar is 16T
/πd³
(b) Determine the angle of twist between the ends of the bar.
let theta () be the angle of twist
polar moment of inertia = πd⁴/32
now from the second image;
lets length dx which is at distance of "x" from "B"
Torque distance x
T(x) = tx
Elemental angle twist = d = T(x)dx / G
so
d = tx.dx / G(πd⁴/32)
d = 32tx.dx / πGd⁴
so total angle of twist will be;
=
=
32tx.dx / πGd⁴
= 32t / πGd⁴
= 32t / πGd⁴ [ L²/2]
= 16tL² / πGd⁴
Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴
