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1 year 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 <u>deposition of sediment. </u>

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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Joaquin mows the lawn at his grandmother’s home during the summer months. Joaquin measured the

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a. Direction from east to west, given distance 11.5 meters, and time of 7.10 s

V = displacement/time V = -11.5/7.10 S V = -1.62 m/s (going left)

b. Joaquin reaches his original position. Displacement is now zero.

Velocity of the lawnmower is equal to "zero" but if we calculate for the average speed of the lawn, you just have to add the distance covered and the time it take to go back at the original position or point of origin

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2 years ago

A speaker vibrates at a frequency of 200 hz. What is it's period ?

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200 Hz = 200 cycles per sec

<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>

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3 years ago

All ions are atoms with a

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All ions are atoms with a charge

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3 years ago

A light ray transfers from more to less dense medium at certain condition , if we repeat this experiment by increasing the angle

evablogger [386]

Answer:

The correct option is;

Still constant

Explanation:

The relative refractive index ₁n₂ between the two medium can be as follows;

$_1n_2 = \dfrac{The \ speed \ of \ light \ in \ medium \ 1}{The \ speed \ of \ light \ in \ medium \ 2} = \dfrac{c_1}{c_2}$

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.

4 0

2 years ago

A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at re

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Answer:

(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s

(b) the kinetic energy of the bullet plus the block before the collision is 500J

Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

Part (A)

Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the block after the bullet embeds itself in the block

(0.1 x 100) + (3 x 0) = v (0.1 + 3)

10 = 3.1v

v = 10/3.1

v = 3.226 m/s

Part (B)

Initial Kinetic energy

Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

Part (C)

Final kinetic energy

Kf = ¹/₂m₁v² + ¹/₂m₂v²

Kf = ¹/₂v²(m₁ + m₂)

Kf = ¹/₂ x 3.226²(0.1 + 3)

Kf = ¹/₂ x 3.226²(3.1)

Kf = 16.13 J

6 0

3 years ago