All categories

3 years ago

Take a long look at the moon. What evidence do you see that indicates that it is NOT tectonically active Explain.

1 answer:

eduard3 years ago

4 0

<span>Without plate tectonics, new mountain belts could not form. Earth would be a Waterworld with occasional shield volcanoes emerging briefly above the waves. If regular catastrophic convective overturn occurred, as on Venus, life would have a precarious foothold indeed.

</span>

You might be interested in

I need help pleaseee

Ludmilka [50]

Answer:

8) 709.8875 J

9) The object is at 7.24375 m from the ground

10) Kinetic energy increases as the object falls.

Explanation:

We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:

h(t) = hi - (g/2) t^2

hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):

h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m

This is the position of the 10 kg object after 0.75 seconds (answer for part 9)

Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:

U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)

Part 10)

the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.

3 0

3 years ago

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