All categories

3 years ago

How does Earth's surface and the structures on the surface change as a result of an earthquake?

Physics

1 answer:

olga2289 [7]3 years ago

8 0

THE CHANGE IN EARTH SURFACE IS DUE TO VIBRATION AND COLLISION IN EARTH TECTONIC PLATES

Explanation:

VIBRATION is the change in the stable state/equilibrium state caused by increased pressure in a particular object or a surface due to electro magnetic wave
COLLISION is the meeting point of two or more surfaces or the objects
During earthquake the tectonic plates of the earth surface begins to collide or cause any irregular drastic movements
This is due to the vibration caused by earthquake
This causes the shake in the upper surface which results in construction damage and calamities
This causes drastic change in the surface of the earth.

You might be interested in

Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I

denis23 [38]

Answer:

$I = I_o (10^{\frac{\beta}{10}})$

Explanation:

As we know that sound level is given as

$\beta = 10 Log (\frac{I}{I_o})$

so here we have

$\frac{\beta}{10} = Log(\frac{I}{I_o})$

now we can take anti log both sides

$10^{\frac{\beta}{10}} = \frac{I}{I_o}$

so we have

$I = I_o (10^{\frac{\beta}{10}})$

so above is the expression for the intensity of sound in terms of reference intensity and sound level

4 0

3 years ago

The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con

faust18 [17]

When light travels from a medium with greater refractive index $n_1$ to a medium with smaller refractive index $n_2$ , there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
In this problem, we know that the critical angle is $\theta_c = 53.7^{\circ}$ , so we can find the ratio between the refractive indices of the two mediums:
$\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81$
and since the second medium is air (n=1.00), the refractive index of the first medium is
$n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23$

In the second part of the problem, we have light entering from air ( $n_i = 1.00$ ) at angle of incidence of $\theta_i = 45.0 ^{\circ}$ , into the second medium with $n_r = 1.23$ . By using Snell's law, we can find the angle of refraction of the light inside the medium:
$n_i \sin \theta_i = n_r \sin \theta_r$
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574$
$\theta_r = \arcsin(0.574)=35.1^{\circ}$

3 0

3 years ago

Express 2759 mockingbirds in kilomockingbirds. answer in units of kmockingbird

Nadya [2.5K]

I would think 2.759kmockinbirds if you consider mockingbirds equal to a metre and kmockinbird to kilometres

6 0

3 years ago

What is happening in the first part of this picture?

d1i1m1o1n [39]

Answer:

kenitic energy

Explanation:

6 0

3 years ago

What two things affect the density of Water?

nikdorinn [45]

The temperature of the water and the and the salinity of water

5 0

3 years ago