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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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4 years ago
A concave lens has a focal length of -31 {cm}. Find the image distance and magnification that results when an object is placed 2
kolezko [41]

Answer:

The value is v  =  440 \  cm

and  

 m  =  16

Explanation:

From the question we  are told that

    The image  distance is  v  =  29 \  cm

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From the lens equation we have that

      \frac{1}{f}  =  \frac{1}{v}  - \frac{1}{u}

=>    \frac{1}{-31}  =  \frac{1}{v}  - \frac{1}{29}

=>     v  =  450 \  cm

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=>    m  =  \frac{ 450}{29}

=>    m  =  16

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Answer:

v= 20.8 m/s

Explanation:

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       v_{f} = v_{o} + a*t  =  v_{o} + g*t (1)

  • We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.
  • Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:

        \Delta y = v_{o} *t + \frac{1}{2} * a* t^{2}  (2)

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  • Replacing (4) , and the values of g and t in (1) we can find the value that we are looking for, vf:

       v_{f} = v_{o} + g*t  = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)

  • Therefore, the speed of the ball (the magnitude of the velocity) as it passes the top of the window is 20.8 m/s.
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