The magnitude of the electric field at the third vertex of the triangle is determined as zero.
<h3>Electric field at the third vertex of the triangle </h3>
The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;
E = E(13) + E(23)
E = (kq₁)/r² + (kq₂)/r²
where;
- q1 is positive charge
- q2 is negative charge
E = (kq₁)/r² - (kq₂)/r²
E = 0
Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.
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Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
q = f₁
2) for an object located at p = 25 cm
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
= 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = 
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
<span> U = (1/2)kx^2
</span><span> U = (1/2)(5.3)(3.62-2.60)^2
</span> U = <span>
<span>2.75706 </span></span>J
By definition we know that the force is the vector product of the vector of the current by the length with the magnetic field vector. The current in this case goes in a positive "Y" direction. If we assume that the magnetic field goes in the positive "K" direction, then the result will be in the positive "X" direction. Attached solution.