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jekas [21]
3 years ago
6

Europe and North America are moving apart by about 5 m per century. As the continents separate, new ocean floor is created along

the mid-Atlantic Rift. If the rift is 5000 km long, what is the total area of new ocean floor created in the Atlantic each century?
Physics
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

Area = 25km² in a century.

Explanation:

Given

Spread Rate = 5m/century

Length of rift = 5000km

Convert to metres

Length of rift = 5000 * 1000m

Length of rift = 5,000,000m

In one century, the additional area to Atlantis = (Rift Length) * (Spread rate) * 1 century

Atea = 5,000,000 m * 5m/century * 1 century

Area = 25,000,000m² in a century

----- Convert to km²

Area = 25,000,000km² * 1m²/1,000,000km²

Area = 25km² in a century.

Hence, the total area of new ocean floor created in the Atlantic each century is 25km²

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Given the function f(x) = 8x3 − 3x2 − 5x 8, what part of the function indicates that the left and right ends point in opposite d
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Let the left hand side be donated by -x.

Then,

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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

4 0
3 years ago
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