All categories

3 years ago

What is beach nourishment

1 answer:

4 0

Beach nourishment describes a process by which sediment, usually sand, lost through longshore drift or erosion is replaced from other sources. A wider beach can reduce storm damage to coastal structures by dissipating energy across the surf zone, protecting upland structures and infrastructure from storm surges, tsunamis and unusually high tides

You might be interested in

What color of light must blue light mix with to create white light?

schepotkina [342]

<h2>Answer: a. Yellow </h2>

According to the additive theory of color, when we join the <u>three primary colors of light</u> (Red + Green + Blue) we get White light.

On the other hand we have <u>secondary colors of ligh</u>t that are:

Yellow = Red + Green

Magenta = Blue + Red

Cyan = Blue + Green

Now, if we know that:

Red + Green + Blue = White

And:

Red + Green = Yellow

Then:

<h2>Yellow + Blue = White</h2>

5 0

3 years ago

If the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force

just olya [345]

The object would stay constant

5 0

3 years ago

True or false is a chemical compound made up of just one element ?¿

Rina8888 [55]

False because compound means two so a chemical compound is made up of two chemicals

3 0

3 years ago

Read 2 more answers

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$