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-BARSIC- [3]
3 years ago
5

A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.

Physics
1 answer:
tatiyna3 years ago
5 0

Answer:

a)   x = ⅔ d , b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

         ∑ F = 0

        -F₁₂ + F₂₃ = 0

         F₁₂ = F₂₃

         

let's replace the values

        k Q Q / r₁₂² = k Q 4Q / r₂₃²

            Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

        r₁₂ = x

        r₂₃ = d-x

             

we substitute

             1 / x² = 4 / (d-x) 2

             1 / x = 2 / (d-x)

             x = 2 (x-d)

             x = 2x -2d

            3x = 2d

              x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c)  Q

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A 920-kg compact car moving at 92 m/s has approximately 3,893,440 Joules of kinetic energy. What is the change in kinetic energy
tensa zangetsu [6.8K]

Answer:

Change in kinetic energy = 3297280 J

Explanation:

Given that,

Mass, m = 920 kg

Speed of a car, v = 92 m/s

Kinetic energy, K = 3,893,440 J

If the speed of a car, V = 36 m/s

Net kinetic energy is given by :

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3 0
2 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
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Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

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ma = qVB

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a is acceleration

q is charge

V is velocity

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We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

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Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

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a = -0.33 m/s² k^

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