1. A pendulum is 0.25 m long. What is the frequency of its oscillations? How do you solve this
1 answer:
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Answer:
a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)
b. Apparent weight at the top is approximately 519.06 N
Apparent weight at the bottom is approximately 558.94 N
Explanation:
a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly
b. The given parameters are
The radius of the Ferris wheel, r = 7.2 m
The period for one complete revolution, t = 28 seconds
The angle covered in one revolution, θ = 2·π radian
The mass of the person riding on the Ferris wheel, the passenger = 55 kg
Therefore, we have;
The angular speed, ω = Δθ/Δt = 2·π/(28)
From which we have;
Centripetal force, = m × ω² × r
Substituting the known values, we have = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N
The centripetal force, = 19.94 N always acting outward from the center
Weight = Mass × Acceleration due to gravity
The weight of the passenger = 55 kg × 9.8 m/s² = 539 N
The weight of the passenger = 539 N always acting downwards
At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards
Therefore;
The net force, which is the apparent weight of the passenger at the top = 539 N - 19.94 N ≈ 519.06 N
Apparent weight at the top ≈ 519.06 N
At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards
Therefore;
The net force at the bottom, which is the apparent weight of the passenger at the bottom = 539 N + 19.94 N ≈ 558.94 N
Apparent weight at the bottom ≈ 558.94 N.
Refer to the diagram shown below.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Answer:
a) p=0, b) p=0, c) p= ∞
Explanation:
In quantum mechanics the moment operator is given by
p = - i h’ d φ / dx
h’= h / 2π
We apply this equation to the given wave functions
a) φ =
.d φ dx = i k
We replace
p = h’ k
i i = -1
The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero
p = 0
b) φ = cos kx
p = h’ k sen kx
The average sine function is zero,
p = 0
c) φ =
d φ / dx = -a 2x
.p = i a g ’2x
The average moment is
p = (p₂ + p₁) / 2
p = i a h ’(-∞ + ∞)
p = ∞