Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.
Explanation: To find the answer, we have to know more about the equipotential surfaces.
<h3>What are equipotential surfaces?</h3>
- An equipotential surface is the locus of all points which have the same potential due to the charge distribution.
- Any surface in an electric field, at every point of which, the direction of electric field is normal to the surface can be regarded as equipotential.
- We have the equation for electric potential as,
, where k is equal to 1/(4π∈₀) = .
- equation for radial distance will be,
<h3>How to solve the problem?</h3>
- For the first surface, we can write the equation of potential as,
- For the second surface, we can write the equation of potential as,
- Thus, the radial distance will be,
Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.
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The answer is B.
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Answer:
Kinetic energy is energy possessed by an object in motion. The earth revolving around the sun, you walking down the street, and molecules moving in space all have kinetic energy. Potential energy is energy an object has because of its position relative to some other object.
Explanation:
The force experienced by the foot is 1000 N
Explanation:
We can solve the question by applying Newton's third law, which states that:
<em>"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
</em>
In this situation, we can identify the foot as object A, and the ball as object B.
We are told that the ball experiences a 1000 N force, so the foot exerts a force of 1000 N on the ball (action). As a consequence of Newton's third law, therefore, the ball also exerts an equal and opposite force of 1000 N (reaction) on the foot.
It is important to remember that action and reaction do not act on the same object, so they never appear at the same time in the same free-body diagram (which shows only the forces acting on one object).
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