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Bad White [126]
3 years ago
13

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

a distance 0.0429 m from the origin. Where should a third particle with charge 9.03 µC be placed so that the magnitude of the electric field at the origin is zero?
Physics
1 answer:
ra1l [238]3 years ago
3 0

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

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<h2>Given that,</h2>

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Final speed of car Z, v₂ = 10 m/s

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m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

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Stephanie serves a volleyball from a height of 0.80 m and gives it an initial velocity of +7.2 m/s straight up. how high will th
Papessa [141]
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bonufazy [111]

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If we solve Eq (1) for the velocity (v) we obtain:

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Crazy boy [7]
A) Work energy relation;
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But v = u-at
  0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
6 0
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