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Alisiya [41]
3 years ago
7

Help!!! Need answer ASAP.

Physics
1 answer:
egoroff_w [7]3 years ago
6 0

Answer:

Hey

Explanation:

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To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

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Here,

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From each object the potential will be

V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}

Replacing the values we have that

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}

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Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}

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Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

\frac{1}{2} mv^2 = q(V_2-V_1)

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Rearranging to find the velocity

v = \sqrt{ \frac{2q(V_2-V_1)}{m}}

Replacing,

v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}

v = 6.3*10^6m/s

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is 6.3*10^6m/s

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