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Bumek [7]
3 years ago
5

An electron and a positron are located 15 m away from each other and held fixed by some mechanism. The positron has the same mas

s and the same magnitude of charge as those of the electron, but its charge is positive. The electron and the positron are released at the same time by the mechanism. The electron and the positron begin to speed up towards each other. What velocities should they have when they are 2 m away from each other
Physics
1 answer:
ICE Princess25 [194]3 years ago
8 0

Answer:

Explanation:

Electrical potential energy will be converted into kinetic energy .

Electrical potential energy when distance was 15 m .

E₁ = 9 x 10⁹ x - q² /d where q is magnitude of charge on electron or positron

E₁ = 9 x 10⁹ x - ( 1.6 x 10⁻¹⁹ )² /15

= - 1.536 x 10⁻²⁹ J .

Electrical potential energy when distance was 2 m .

E₁ =9 x 10⁹ x - q² /d where q is magnitude of charge on electron or positron

E₁ = 9 x 10⁹ x - ( 1.6 x 10⁻¹⁹ )² /2

= -11.52 x 10⁻²⁹ J .

Decrease in energy = (11.52 - 1.536 ) x 10⁻²⁹

= 9.984 x 10⁻²⁹ J .

This energy will be converted into kinetic energy and they will be distributed equally in each .

Energy of each = 9.984 x 10⁻²⁹ /2

= 4.992 x 10⁻²⁹ J .

1/2 m v² = 4.992 x 10⁻²⁹ , m is mass of electron

.5 x 9.1 x 10⁻³¹ v² = 4.992 x 10⁻²⁹

v² = 109.71

v = 10.47 m/s .

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Answer:

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Explanation:

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For the first situation:

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(1)X*M=-0,015kgm

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Now the weights are interchanged, and as the pivot point has change, the distances from the pivot point are different:

We have a 0,3kg mass al -0,33m from the pivot point, and a 0,15kg mass at +0,32m of the pivot point. As the pivot has moved we have to move our reference for X: The center of gravity of the stick is at a X+0,07m distance with a mass M, the torque sum:

0,3kg*(-0,33m)+0,15kg*0,32m+(X+0,07m)*M=0

(X+0,07m)*M=0,051kgm

X*M+0,07m*M=0,051kgm

replacing X*M from (1):

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and replacing this in (1):

X*0,9428kg=-0,015kgm

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As we took the 50cm mark as the reference, the center of mass is at the 48,41cm  mark

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Answer:

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