PLEASE HELP 15 POINTS The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. P
lease use 2 content related sentences
1 answer:
The field lines spread apart as we move away from the charge, and they point away from the charge
Explanation:
The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation
where
k is the Coulomb's constant
Q is the magnitude of the charge
r is the distance from the charge at which the field is calculated
There are two pieces of information given by the field lines shown in the graph:
- The spacing between the lines gives an indication of the strength of the field: the closer to each other they are, the stronger the field. In this case, as we move away from the charge, the spacing between the lines increases, and this means that the field becomes weaker (in fact, it follows an inverse square law,
- The direction of the lines gives the direction of the electric field, which points away from the central charge. This is because the direction of the electric field corresponds to the direction of the force that a positive test charge would feel when immersed in the electric field: in this case, if we place a positive test charge in this field, then it would get repelled away from the central charge (remember that the electric force between two positive charges is repulsive), and therefore, the direction of the electric field is away from the central charge.
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Answer:
h=2.86m
Explanation:
In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,
There is no kinetic energy in the initial state, nor potential energy in the end,
In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,
The inertia of the bodies is given by the equation,
On the other hand the angular velocity is given by
Replacing these values in the equation,
Solving for h,
Answer:
The trains will collide at a distance 1660 m from the station
Explanation:
Let the train traveling due north with a constant speed of 100 m/s be Train A.
Let the train traveling due south with a constant speed of 136 m/s be Train B.
From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).
Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.
∴ If train A has covered a distance m by the time of collision, then train B would have traveled
m.
Also,
At the position where the trains will collide, the two trains must have traveled for equal time, t.
That is, At the point of collision,
is the time spent by train A
is the time spent by train B
From,
Since the time spent by the two trains is equal,
Then,
m
m
m/s
m/s
Hence,
≅ 1,221 m
This is the distance covered by train A by the time of collision.
Hence, Train B would have covered (2881 - 1221)m = 1660 m
Train B would have covered 1660 m by the time of collision
Since it is train B that leaves a station,
∴ The trains will collide at a distance 1660 m from the station.