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3 years ago

What is a tsunami? How is it formed?

2 answers:

4 0

A tsunami is a series of waves generated in an ocean or other body of water by a disturbance such as an earthquake, landslide, volcanic eruption, or meteorite impact. ... Undersea earthquakes, which typically occur at boundaries between Earth's tectonic plates, cause the water above to be moved up or down

PtichkaEL [24]3 years ago

4 0

Answers:

1. A tsunami, also known as a seaquake is a seism whose epicenter is located at the depths of the sea or ocean (or any a mass of water) producing the agitation of the waters and forming large waves.

2. How is it formed?

After the seism at the depth of the mass of water begins, and if the seism has a vertical component, the movement causes a displacement of the sea bottom and, consequently, it causes the formation giant waves as the water is displaced because of gravity, trying to return to a stable position.

Now, when the tsunami approaches the coast its speed decreases, because the depth decreases, in other words, the water accumulates when braking, increasing the height of the wave.

And, according to the conservation of energy principle, if the speed of the wave decreases, its height increases.

You might be interested in

Geologists use the blank test to determine the hardness of a mineral

Ludmilka [50]

Geologists use the Mohs Harness Scale (aka the Scratch Test) to determine the hardness of a mineral.

For more info, check out http://geology.com/minerals/mohs-hardness-scale.shtml

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3 years ago

A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same

emmainna [20.7K]

This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal.
momentum = mass x velocity
(1,600 kg)(16 m/s) + (1.0x10^3 kg)(10 m/s) = (1600 + 1000 kg)(x)
The value of x is 13.69 m/s. Thus, their final speed is approximately letter D. 14 m/s.

6 0

4 years ago

A long distance runner sees the finish line and accelerates at a rate in 1.2 m/s2 for

Nuetrik [128]

Answer: he did travel 15 meters.

Explanation:

We have the data:

Acceleration = a = 1.2 m/s^2

Time lapes = 3 seconds

Initial speed = 3.2 m/s.

Then we start writing the acceleration:

a(t) = 1.2 m/s^2

now for the velocity, we integrate over time:

v(t) = (1.2 m/s^2)*t + v0

with v0 = 3.2 m/s

v(t) = (1.2 m/s^2)*t + 3.2 m/s

For the position, we integrate again.

p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0

Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m

Then the displacement at t = 3s will be equal to p(3s).

p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m

6 0

3 years ago

What is the acceleration of a 1000 kg car subject to a 550 N net force

Xelga [282]

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

PLEASE MARK ME AS BRAINLIEST!!!

4 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

3 years ago