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Reika [66]
3 years ago
6

An electron and a positron are located 17 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 1.3 m away from each other?

Physics
1 answer:
mariarad [96]3 years ago
6 0

Answer:

Velocity will be 13.9 m/s when they are 1.3 m away from each other.

Explanation:

Detailed steps are attached below.

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Ad libitum [116K]

Answer:

h =220 m

Explanation:

Given that

u = 7 m/s

Even mass will attach but this will not produce any effect on the maximum height of the ball.Because in energy conservation the effect of mass does not present.

So the final speed of the ball will be zero at the maximum height.

v² = u² - 2 g (25 + h)

0 = 7² - 2 x 10 (25 +h)

49 = 20 ( 25 +h)

49 = 500 +20 h

Here h comes out negative that is why we are taking the 70 m/s in place of 7 m/s.

0 = 70² - 2 x 10 (25 +h)         ( take g =10 m/s²)

4900 = 20 ( 25 +h)

4900 = 500 +20 h

4900- 500 = 20 h

4400 = 20 h

440 = 2 h

h =220 m

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3 years ago
How can you demonstrate that charged objects exert forces, both attractive and repulsive
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If it attractive it has opposite pole and if it repulsive it has same pole
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A care package is dropped from an airplane flying at 1000 m above sea level with a velocity of 200 m/s to an island. If it takes
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Answer:3000

Explanation:

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How does Physics help you as a student?
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Answer:

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Read 2 more answers
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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