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AnnyKZ [126]
3 years ago
9

Point P and point charge Q are separated by a distance R. The electric field at point P has magnitude E. How could the magnitude

of the electric field at point P be doubled to 2E?,
Physics
1 answer:
Angelina_Jolie [31]3 years ago
5 0
<span>We know , E = kQ/r^2 where q = charge and r is separation between point and point charge. Now, At P, E= kQ/r^2 Since, Q can't be changed, we can do that by varying r 2E = 2kq/r^2 2E = kq/ (r/ sqrt2)^2 Hence, if we bring Q closer such that distance between P and Q becomes r/ sqrt 2, E will get doubled.</span>
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(a) Write the energy equation for an elastic collision. (b) For an inelastic collision.
velikii [3]

Answer:

Explanation:

There are two types of collision.

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The kinetic energy of the system before collision = the kinetic energy after the collision

(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.

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4 0
3 years ago
WHAT THE HECK PLEASE HELP HURRY
Angelina_Jolie [31]

i dont know about this but im willing to help you even im a questioneer....

<h2>last choice:blocks 1,2 and 3</h2>

<h3>-questioneer </h3>

comment down below if you know anything about this and explain me so that i can answer it properly

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