If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will attract each other.
<h3>What is electric current?</h3>
Electric current can be defined as the flow of electrons.
Since electrons are easily removed from atom and are very mobile, the flow of electrons constitute an electric current.
Materials which allow electric current to flow through them are known as conductors. Examples of conductors are metals, and electrolytes.
On the other hand, materials which do not allow electric current to pass through them are known as insulators. Examples of insulators are wood and rubber.
The flow of current is known as electricity.
Parallel conductors with current flowing through them in the same direction are attracted to each other as a result of a magnetic field produced by the flow of current.
In conclusion, conductors allow electric current to pass through and the flow of current through a conductor produces a magnetic field.
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The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:
d = vt = (22 m/s)(12 s) = 264 m
For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²
Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m
Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.
There are several information's of immense importance already given in the question. Based on the given information's the answer to the question can easily be determined.
Distance covered by the bicycle = 5000 meter
Time taken by the bicycle to reach the distance = 500 second.
Velocity of the bicycle = Distance / Time taken
= 5000/500 meter/second
= 50 meter/second
So the velocity of the bicycle is 50 meter per second. I hope the procedure is clear enough for you to understand. In future you can always use this procedure for solving similar problems.
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