The vertical height of an ocean wave from a crest to a trough is 10 meters. What is the amplitude of the ocean wave?
2 answers:
Answer:
Amplitude, A = 5 m
Explanation:
It is given that, the vertical height of an ocean from a crest to trough is 10 meters.
The attached figure shows the vertical height from a crest to a trough is 10 m. We know that the maximum displacement of the wave is called its amplitude. A represents the amplitude of wave. It is the distance from the baseline to the crest or to the trough.
So, the amplitude of the ocean wave must be equal to half of wave's height i.e. 10/2 = 5 meters.
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Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
1 / p = 0.107375
p = 93.13 cm
Answer:
a) w = 2.57 rad / s , b) α = 3.3 rad / s²
Explanation:
a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground
Initial. Higher
Em₀ = U = m g h
Final. Touching the ground
= K = ½ I w²
How energy is conserved
Em₀ =
mg h = ½ I w2
The moment of specific object inertia
I = m L²
We replace
m g h = ½ (mL²) w²
w² = 2g h / L²
The height of the object is the length of the bar
h = L
w = √ 2g / L
w = √ (2 9.8 / 2.97)
w = 2.57 rad / s
b) the angular acceleration can be found from Newton's second rotational law
τ = I α
W L = I α
mg L = (m L²) α
α = g / L
α = 9.8 / 2.97
α = 3.3 rad / s²