Keisha writes that if an object has any external forces acting on it, then the object can be in dynamic equilibrium but not static equilibrium.

which statement best describes keisha’s error?
a)an object that is not moving is always in static equilibrium.
b)an object that is moving must be in dynamic equilibrium.
c)an object in either state of equilibrium must have no forces acting on it.
d)an object in either state of equilibrium must have no net force acting on it.

Before coming to the conclusion,first we have to know about the static and dynamic equilibrium.

As per Newton's first law ,a body continues to be at state of rest or of uniform motion along  a straight line as long as there is no external unbalanced force acting on it.

Hence if the net force is zero,the body  at rest will be at rest.This is called static equilibrium .Hence the condition for a body to be in static equilibrium is that sum of mechanical force is zero.

i.e

Again if the net force is zero,then velocity of the body won't be changed.Hence the acceleration of the body will be zero.This is called dynamic equilibrium where a body will move with uniform velocity  in the absence of net external force.

Hence if the net force is zero,then body will be both in static and dynamic equilibrium.

Hence the option  D  is right which  tells that a body in either state of equilibrium has no net force acting on it.

The answer  is D. on Edgen.

Explanation:

D. An object in either state of equilibrium must have no net force acting on it.

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

c) up is the direction the magnetic force is acting on the charge.