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2 years ago

when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results

1 answer:

4 0

When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results deposition of sediment.

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

Hard stabilization is the prevention of erosion through the use of artificial barriers.
Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.

Learn more about the Hard stabilization with the help of the given link:

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3 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

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Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

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3 years ago

A race car starting from rest accelerates uniformly at a rare of 4.90 meters per second^2. What is the cars speed after it has t

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From the law of Galileo Galilei :v²=v₀²+2ad we take the speed

v²=0+2*4.90*200=1960=>v=√1960=44.27 m/s

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3 years ago

Conditional probability involves events that are independent. true or false?

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Its false I got It right .

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3 years ago

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

FrozenT [24]

Answer: Option (C) is the correct answer.

Explanation:

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Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

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3 years ago

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