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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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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

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Numerical Problems:
dangina [55]
  • Displacement=1200m
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\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

\\ \sf\longmapsto Velocity=\dfrac{1200}{240}

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In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init
liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,  

n  = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,  

U = − 0.061 J

The charge on an electron,  

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,  

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:  

q ₁ = n q

   = 1 ×10 ¹³ *  − 1.6 × 10 ⁻¹⁹

   = -1.6  × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6  × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and  q ₂ are the two charges.

r  is the distance between the charge and the point.

K  =  8.98755  ×  10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 =  (8.98755  ×  10 ⁹ * (-1.6  × 10⁻⁶)²) / r

r = (18.41 ×  10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0
3 years ago
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