Pitch is way to relate a sound to its frequency. High frequencies have high pitches (think of a flute), and low frequencies have low pitches (think of a bass). <span>
</span>
Answer:
1.5 kgms⁻¹
Explanation:
Momentum can be defined as "<em>mass in motion</em>."
The amount of momentum that an object has is dependent upon two factors
- mass of the moving object
when there is a change in the velocity , it creates a change in momentum also
when we consider that we can mathematically show this,In terms of an equation,
Change in momentum (ΔΡ) = m(Δv)
where (Δv) - change in velocity
<em>(Δv) = final velocity - initial velocity</em>
Change in momentum (ΔΡ) = m(Δv)
= 0.1×([55-40])
= 1.5 kgms⁻¹
Alnilam is the brightest star
Absolute brightness is how bright the star actually is. Apparent brightness is how bright the star looks.
Factors that influence how bright a star looks is distance. The fact that Alnilam has the same apparent brightness of the other two stars even though it is farther away, means that it is brighter.
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