Answer: The same current flows through bth cables
Explanation:
Lets have a look to the next two equations
The Ohm´s V = I*R (1)
where:
V is voltage (potencial dfference) in volts
I is the electric current in ampers
R is the electric resistance
When a voltage is applied as the electrc load is not specified ( we have to assume is the same) the current will be the same
And in the other hand the resistance R =ρL/s
Where ρ is the resistivity of the conductor L the length and s square section of the conductor
If we assume that the smaller diameter cable is able to conduct the current then nothing happens. The point is that the capacity of conduction of current depend on the section of the cable (the area)
Tables exist where to find the capacity of each cable according to its diameter.
The correct answer to the question is- 
CALCULATION:
As per the question, the electric field generated by the source charge is 1236 N/C at a distance of 4 m.
Hence , electric field E = 1236 N/C.
The distance of the point R = 4m
We are asked to calculate the charge possessed by the source.
The electric field produced by a source charge of Q at a distance R is calculated as -
Electric field E = 
Here,
is called the absolute permittivity of the free space.
Hence, the charge of source is calculated as -
Q = 
= 
= 
= 
= 
Hence, the charge of source is 
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .