What are the three types of friction and when does each apply
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:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 × m
diameter d = 0.074 cm = 74 × m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W