All categories

2 years ago

What causes coastal erosion

1 answer:

5 0

La erosión costera es la pérdida o desplazamiento de tierra, o la remoción a largo plazo de sedimentos y rocas a lo largo de la costa debido a la acción de olas, corrientes, mareas, agua impulsada por el viento, hielo transportado por el agua u otros impactos de marejadas ciclónicas.

You might be interested in

Which of these was Sir Isaac Newton not known for? invention of calculus invention of the microscope the Three Laws of Motion un

Akimi4 [234]

<span>The correct answer is option B. i.e.invention of the microscope. Galileo Galilei invented the microscope and showed its use. Sir Issac Newton invented some part of the calculus, Also invented Newton's Laws of motion and the universal gravitation theory.</span>

3 0

3 years ago

Read 2 more answers

1. A man travels 45 m in 35 seconds. How fast is that?<br> What’s the answer

Natalija [7]

Answer:

1.28

Explanation:

If you want to find the m/s you would divide distance by time, so

45 divided by 35 would equal 1.28571429 and so on.

you can just write the three first numbers.

8 0

3 years ago

What is the acceleration of the the object during the first 4 seconds?

AVprozaik [17]

Answer:

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

6 0

2 years ago

I don’t understand I need help asap

hammer [34]

For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s

3 0

2 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago