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sp2606 [1]
3 years ago
9

Two charges, q1 and q2, are separated by a certain distance r. if the magnitudes of the charges are halved and their separation

is halved, then what happens to the electrical force between these two charges?
Physics
1 answer:
vazorg [7]3 years ago
6 0

The electrical force between these two charges remains the same. In coulomb’s law, it states that the magnitude of two charges (product of two charges) is inversely proportional to the square of the distance. Since both the magnitude and the distance are halved, therefore, the change in both quantities will have no effect in the value of electrical force.

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A volcano erupts spewing ash into the air and sending lava flowing down the side of the mountain. Looking at the image explain h
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a)       a = 3.72 m / s², b)    a = -18.75 m / s²

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as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

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