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Natali [406]
3 years ago
15

A electron is released from rest in a uniform electric field oriented from left to right. What happens to the electric potential

energy of the proton-electric field system?A. IncreaseB. DecreaseC. Remains the same
Physics
1 answer:
True [87]3 years ago
4 0

Answer:

Explanation:

When electron is released in an electric field , it will move in the direction opposite to the direction of field. In this process , force is applied by the field and work is done by the field. So its electric potential energy will be decreased because , its energy is converted into kinetic energy of the electron .

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frez [133]
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3 years ago
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grandymaker [24]

Answer:

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Explanation:

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4 0
3 years ago
What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?
sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

F=k\dfrac{q_1q_2}{d^2}

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F

So, the new force becomes 4 times the initial force.

4 0
3 years ago
A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image dista
KonstantinChe [14]

Answer:

-353.16

Explanation:

4 0
3 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

4 0
3 years ago
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