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Anestetic [448]
4 years ago
7

an electron is held at a distance from a stationary, positively charged object . If the electron is released , what energy chang

e will take place?
Physics
2 answers:
defon4 years ago
5 0
Electric potential energy will be transformed into kinetic energy.
34kurt4 years ago
5 0

Answer:

Electrostatic potential energy will convert into kinetic energy

Explanation:

When electron is released from rest at some distance from positive charge then due to electrostatic attraction force these two charges will attract each other.

Now in the form of energy we can say that

electrostatic potential energy = \frac{kq_1q_2}{r}

now as the two charges comes closer to each other then the electrostatic potential energy will decrease and hence this energy will convert into kinetic energy of two charges

so we can say

\frac{kq_1q_2}{r_2} - \frac{kq_1q_2}{r_1} = \frac{1}{2}mv^2

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A wave in which particles of the medium move at right angles to the direction of the wave is called a
Licemer1 [7]

Answer:

transverse wave: A wave in which particles of the medium move at right angles to the direction of the wave is called a transverse wave.

5 0
3 years ago
At which type of boundary between two tectonic plates is an earthquake or a tsunami most likely to form?
Fofino [41]

Answer:

B. convergent boundary

Explanation:

Convergent boundaries, where ocean plates are subducted at trenches, produce the most tsunamis because they produce the largest earthquakes

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3 years ago
Please help with these i dont know how to do them
Svetradugi [14.3K]

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3 0
4 years ago
An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t
scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

C = \frac{Q}{V}  = e_0\frac{A}{d}

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

V = \frac{Q}{e_0} *\frac{d}{A}

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V

So, if we double the distance between the plates, the potential difference will also be doubled.

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