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Karo-lina-s [1.5K]
3 years ago
9

Formed through longshore drift a. sea stack b. sandbar c. spit d. headland

Physics
2 answers:
ANTONII [103]3 years ago
7 0
<h3>Answer;</h3>

<em><u>Sand Spit or Spit </u></em>

<h3><u>Explanation;</u></h3>
  • <em><u>Long shore drift is the process that occurs when a sheet of water moves on and off the beach, in other words the swash and back swash</u></em>, thus capturing and transporting sediment on the beach back out to the sea.
  • <em><u>Sandbar</u></em> is normally formed when the sandspit stretches across a bay and connects the two sides. <em><u>Headland</u></em> is a high piece of land that extends out onto the sea. <em><u>Sea stacks </u></em>on the other hand results from the collapsing of the roof of the arch.
pochemuha3 years ago
5 0

<u>Option (c) is correct. The sand spits or the spits are formed by the long shore drifts that flow along the shore.</u>

Explanation:

The long shore drifts are the waves or the geological process that move obliquely along the shore of the coast and are responsible for carrying the sediments like the sand, silt, clay etc. to the shore.

The long and stretched deposition of the sediment by the long shore drifts carrying the sediments lead to the formation of the sand spits or simply the spits along the shores of the coast.

The sand are very similar to the deposition of the spits but they are formed where the two ends connects the bay. The sea stack on the other hand are the unique structures formed in the sea that is formed in the sea near the coast and it is mostly made up of the rock.

The headland  is a very high region near the coast with a sudden or a very steep slope.

Thus, <u>Option (c) is correct. The sand spits or the spits are formed by the long shore drifts that flow along the shore.</u>

<u></u>

Learn More:

1. The universal force that is effective over long distance is brainly.com/question/2957659

2. The type of mirror do dentist use brainly.com/question/997618

3. Forces of attraction limit the motion of particles most in brainly.com/question/947434

Answer Details:

Grade: High School

Subject: Physics

Chapter: Sediments

Keywords:

long shore, drift geological, waves, sediments, sand, sandbar, headland, spit, sea stack, structure, shore, coast.

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Likurg_2 [28]

Answer:

you can't

Explanation:

you can only comment on your question and add more details or can ask a moderator or administrator to remove your question with a reason

3 0
2 years ago
A thin, horizontal copper rod is 1.0792 m long and has a mass of 53.1794 g. The acceleration of gravity is 9.8 m/s 2 . What is t
horsena [70]

Answer:

0.1675 A

Explanation:

Since BILsin\theta=mg then making I the subject I=\frac {mg}{BLsin\theta} where L is the length of the rod, in this case given as 1.0792 m, B is magnetic field which is given as 2.88578 T, m is the mass which is  53.1794 g which is equivalent to 0.0531794 Kg . For minimum current, sin\theta=1.

Substituting the given values then

I=\frac {0.0531794\times 9.81}{1.0792\times 2.88578}=0.167512525 A\approx 0.1675 A

3 0
3 years ago
Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another
Tanya [424]

Answer:I_2=0.618\times 10^{-7} W/m^2

Explanation:

Given

Distance between source and  receiver d_1=3 m

Sound Intensity I_1=1.1\times 10^{-7} W/m^2

Distance of of second observer d_2=4 m

Intensity varies as

I\propto \frac{1}{d^2}

using this

I=\frac{k}{d^2}

\frac{I_1}{I_2}=\frac{d_2^2}{d_1^2}

\frac{1.1\times 10^{-7}}{I_2}=\frac{4^2}{3^2}

I_2=0.75^2\times 1.1\times 10^{-7}

I_2=0.618\times 10^{-7} W/m^2

     

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2 years ago
A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and a
NikAS [45]

Answer:

a=1.024m/s

t=15.62s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t                         (1)

{Vf^{2}-Vo^2}/{2.a} =X      (2)

X=Xo+ VoT+0.5at^{2}      (3)

X=(Vf+Vo)T/2                   (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this problem

Vf=16m/s

Vo=0m/s, the cart starts from the rest

X=125m

we can use the ecuation number tow to calculate the acceleration

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{16^{2}-0^2}/{2(125)} =a

a=1.024m/s

to calculate the time we can use the ecuation number 1

Vf=Vo+a.t    

t=(Vf-Vo)/a

t=(16-0)/1.024

t=15.62s

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