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MArishka [77]
3 years ago
8

The vibration produced days or even years before an earthquake

Physics
1 answer:
Margaret [11]3 years ago
7 0
I don't know I guess its the plate tectonics 
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Knowledge and skills learned through socialization are an example of
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I think no.2 the answer

Because socialization and social resources are both for me

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Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
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2.521 (A); 14.0924 (V)

Explanation:

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Which one of the following substances is a liquid fuel used in rocket engines?
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There are none on the list you included with your question.

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A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:
AfilCa [17]

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

Learn more here: brainly.com/question/19072061

8 0
3 years ago
A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally
Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, g=-9.8 m/s^2, downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

v_x = +2.60 m/s

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

v_y = u_y +gt

where

u_y = 0 is the initial vertical velocity, which is zero

g=-9.8 m/s^2 is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

v_y = 0+(-9.8)(1.75)=-17.2 m/s

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0
3 years ago
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