This picture represents the electric field diagram between two particles with static charges. Do the two particles have the same
charge? If the particles are free to move, what affect will the two particles have on each other? If a third particle was added to this field with a positive charge, what would it be attracted to? If an uncharged particle was placed between A and B, which way would it move?
1 answer:
Answers:
No, They will attract each other, B, and neither direction
Explanation:
Since the two already presented particles in the diagram represent both opposing charges due to the direction of the arrows (the arrows facing away from the particle shows a positive charge and the particles facing towards the particle show a negative charge), not only because of this but as the arrows between the particles show an attracting magnetic field, then it can be concluded that the particles will attract to each other and if another particle was introduced into the diagram of a positive charge, then it would attract to the negatively charged particle. If you have any questions or need further explanation, please comment below. E2021, have a great day.
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(a) 3.58 km 45° south of east
The total displacement is given by:
where
v is the average velocity
t is the time
The average velocity is:
v = 3.53 m/s
While we need to convert the time from minutes to seconds:
Therefore, the magnitude of the displacement is
And the direction is the same as the velocity, therefore 45° south of east.
(b) 5.53 m/s 90° south of east
The velocity of the air relative to the ground is
and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is
v = 3.53 m/s
So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed (). Since these two vectors are in opposite direction, we have
Therefore we find v', Allen's velocity relative to the air:
The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.
(c) 56.1 km at 90° south of east.
Since Allen's velocity relative to the air is
v' = 5.53 m/s
Then the displacement of Allen relative to the air will be given by
and substituting,
And the direction is the same as that of the velocity, therefore will be 90° south of east.