A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
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Answer:
The magnitude of the electric field at a point equidistant from the lines is
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.
Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes
Therefore, the current flowing in the first device is 0.033 Amperes or 33 milliAmperes.
Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes
Therefore, the current flowing in the first device is 0.033 Amperes or 33 milliAmperes.