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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 
You might be interested in
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
2 years ago
Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

4 0
3 years ago
Parachutes utilize maximum air ____ in order to limit the acceleration of the fall.
eduard

Answer: Resistance

Explanation:

8 0
3 years ago
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Researchers conducted an experiment to identify the effects of three different hand sanitizers on the growth of bacteria. The re
jek_recluse [69]
The best and most correct answer among the choices provided by your question is the second choice or letter B.

The researchers’ conclusion was not justified because t<span>he control group was not treated the same as the experimental group.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
3 years ago
2. The ability to keep<br> upright posture while standing or moving
Masteriza [31]
Is balancing .
Hope this is helpful
4 0
3 years ago
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