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Soloha48 [4]
3 years ago
6

How does the magnitude of the electrical force compare between a pair of charged particles when they are brought to half their o

riginal distance of separation? To one-quarter their original distance? To four times their original distance? (What law guides your answers?)
Physics
1 answer:
tatiyna3 years ago
5 0

Answer:

a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law

Explanation:

Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.

F= k q1q2 / (r12)2

a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:

F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)

b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)

c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.

The empirical law that allows to find out easily these values, is the Coulomb´s Law.

Explanation:

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21.-Una esquiadora olímpica que baja a 25m/s por una pendiente a 20o encuentra una región de nieve húmeda de coeficiente de fric
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Answer:

y = 12.82 m

Explanation:

We can solve this exercise using the energy work theorem

          W = ΔEm

friction force work is

          W = fr . s = fr s cos θ

the friction force opposes the movement, therefore the angle is 180º

           W = - fr s

we write Newton's second law, where we use a reference frame with one axis parallel to the plane and the other perpendicular

           N -Wy = 0

           N = mg cos θ

the friction force remains

            fr = μ N

            fr = μ mg cos θ

             

work gives

           W = - μ mg s cos θ

initial energy

           Em₀ = ½ m v²

the final energy is zero, because it stops

we substitute

          - μ m g s cos θ = 0 - ½ m v²

          s = ½ v² / (μ g cos θ)

         

let's calculate

          s = ½ 20² / (0.55 9.8 cos 20)

          s = 39.49 m

this is the distance it travels along the plane, to find the vertical distance let's use trigonometry

            sin 20 = y / s

           y = s sin 20

           y = 37.49 sin 20

           y = 12.82 m

8 0
4 years ago
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