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just olya [345]
3 years ago
9

Mendeleev

Physics
2 answers:
Evgesh-ka [11]3 years ago
8 0

Answer:

1

Explanation:

deff fn [24]3 years ago
8 0

Answer:

Dmitri Mendeleev, Russian in full Dmitry Ivanovich Mendeleyev, (born January 27 (February 8, New Style), 1834, Tobolsk, Siberia, Russian Empire—died January 20 (February 2), 1907, St. ... Petersburg, Russia), Russian chemist who developed the periodic classification of the elements

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Elements from opposite sides of the periodic table tend to form ________.
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3 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
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Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

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