Answer:
the second one
Explanation:
When a free positive charge q is accelerated by an electric field, such as shown in Figure 1, it is given kinetic energy. The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let us explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electric potential energy.
The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy (because it depends only on position) than to calculate the work directly.
Answer:
Wave-cut cliff, sea arch, sea stacks
Explanation:
The effect of a wave erosion is made obvious by the structures formed by the wave action.
The high areas of land adjacent to the incoming wave develop the early features or the formation of wave action, which includes the <em>wave-cut cliff</em>
The continuous undercutting of the cliff by the wave results in the formation of the <em>wave cut platform</em>
The effect of the wave further on a cliff, results in the formation of a sea arch and finally a <em>sea stack</em>
Therefore, the correct sequence is the wave-cut cliff, sea arch, sea stacks
Answer:
the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.
hope this helps
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Explanation:
Explanation:
It is given that,
Distance, r = 3.5 m
Electric field due to an infinite wall of charges, E = 125 N/C
We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :
It is clear that the electric field is inversely proportional to the distance. So,
E' = 291.67 N/C
So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.
