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3 years ago

Why is the fact that earth was molten liquid in its early stages the most important event in history

Physics

1 answer:

trasher [3.6K]3 years ago

3 0

It was important because it tells why the denser material had sunken and the lighter is floating on the surface.

<u>Explanation:</u>

Eventually, after around 500 million years, our young planet's temperature warmed to the liquefying purpose of iron—about 1,538° Celsius (2,800° Fahrenheit). This crucial crossroads in Earth's history is known as the iron fiasco. The iron fiasco permitted more prominent, progressively quick development of Earth's liquid, rough material.

We also know that at one point all the material in the planet was molten because the denser material had sunk and the lighter was floating.

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3 years ago

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t

viktelen [127]

Answer:

a) F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ), b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

F = $k \frac{q_2q_2}{r^2}$

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron r₁ = x

right side -electron r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

F_net = k q Q ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$ )

F _net = kqQ ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

let's substitute the values

F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given

x (m) F (N)

0.05 27.0 10-16

0.10 8.10 10-16

0.15 5.76 10-16

0.20 8.10 10-16

0.25 27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m