Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
Answer:
d. the force is reduced by one quarter to 3.6 n
Explanation:
A galvanic cell is formed when two metals are immersed in solu- tions differing in concentration 1 when two different metals are immersed.
<h3>What is galvanic cell?</h3>
- The galvanic cell utilizes the ability to split the flow of electrons in the process of oxidization and reduction, compelling a half-reaction and connecting each with a wire so that a way can be formed for the flow of electrons via such wire.
- A galvanic cell is an electrochemical cell that transforms the chemical energy of a spontaneous redox response into electrical energy. It has an electrical possibility equal to 1.1 V. In galvanic cells, oxidation occurs at the anode and it is a negative plate. Lessening occurs at the cathode and it is a positive plate.
- A galvanic cell is an electrochemical cell that converts the free liveliness of a chemical method into electrical energy. A photogalvanic cell generates species photochemically which react resulting in an electrical current via an external circuit.
To learn more about galvanic cell, refer to:
brainly.com/question/13031093
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C
Because in my opinion they do bending of a light wave as it passes at an angle from one medium to another.
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
