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2 years ago

Which explanation is least likely to be evidence for geological processes changing Earth's surface?

2 answers:

8 0

Aunque la interacción entre Placas Tectónicas es la principal causa de los sismos no es la única. Cualquier proceso que pueda lograr grandes concentraciones de energía en las rocas puede generar sismos cuyo tamaño dependerá, entre otros factores, de qué tan grande sea la zona de concentración del esfuerzo. Las causas más generales se pueden enumeran según su orden de importancia en:

TECTÓNICA: son los sismos que se originan por el desplazamiento de las placas tectónicas que conforman la corteza, afectan grandes extensiones y es la causa que más genera sismos.

VOLCÁNICA: es poco frecuente; cuando la erupción es violenta genera grandes sacudidas que afectan sobre todo a los lugares cercanos, pero a pesar de ello su campo de acción es reducido en comparación con los de origen tectónico.

HUNDIMIENTO: cuando al interior de la corteza se ha producido la acción erosiva de las aguas subterráneas, va dejando un vacío, el cual termina por ceder ante el peso de la parte superior. Es esta caída que genera vibraciones conocidas como sismos. Su ocurrencia es poco frecuente y de poca extensión.

DESLIZAMIENTOS: el propio peso de las montañas es una fuerza enorme que tiende a aplanarlas y que puede producir sismos al ocasionar deslizamientos a lo largo de fallas, pero generalmente no son de gran magnitud.

EXPLOSIONES ATÓMICAS: realizadas por el ser humano y que al parecer tienen una relación con los movimientos sísmicos.

Explanation:

espero que te ayude <3

horsena [70]2 years ago

3 0

Pollution isn’t a geological process.

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2 years ago

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floo

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a) v = 0

b) The aceleration is 1.41 $m/s^{2}$

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

$\sum{F} = 0 = F_{spring} - F_{external}$

$F_{spring} = F_{external}$

$Kx = F_{external}$

$x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm$

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

$\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma$

For the y axis:

$F_N = W = mg = 3*9.8 = 29.4N$

To calculate the force of friction:

$F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N$

Therefore for x axis:

${Kx-F_{external}-F_{friction}}=ma$

$a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}$

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