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Solnce55 [7]
3 years ago
12

Rogue waves are created along the "Wild Coast" off the southeast coast of ________, where the Agulhas Current flows directly aga

inst large Antarctic storm waves.
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0

Answer:

Africa

Explanation:

A rogue wave refers to the wave that is twice the height of a significant wave occurring in a particular area. The significant wave height is generally referred to as the mean of the largest one-third of waves existing at a particular time period. In simple words, a rogue wave is much larger than any other waves that occur at the proximity of the same time.

This rough wave describes the interaction between the ocean and sea current and swelling of waves. It takes place when the large swells in the ocean, also known as the Antarctic storms, strikes with the rapidly traveling Agulhas current, and the curved water current focuses on the energy of the waves.

Thus, these Rogue waves are often generated along the southeastern coastal regions of Africa, where there occurs the convergence of Antarctic storm waves and Agulhas Current.

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It considered as Zero Gage pressure. 
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Which are disadvantages of series circuits? Check all that apply.
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C and F

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A break in one wire causes all current to stop.

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5 0
3 years ago
Read 2 more answers
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
Sergeeva-Olga [200]

Answer:

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

5 0
3 years ago
In deep space, sphere A of mass MA is located at the origin of an x axis and sphere B of mass MB is located on the axis at a poi
Nookie1986 [14]

Answer:

The gravitational potential energy of the two-sphere system just as B is released is

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Explanation:

The gravitational potential energy of two masses (m and M), separated by a distance, d, is given as

U = -(GMm/d)

For our question,

Mass of object 1 = MA

Mass of object 2 = MB

Distance between them = x₁

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Hope this Helps!!!

6 0
3 years ago
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