If 1.0 joule of work is required to move a charge of 1.0 coulomb between two points is an electric field the potential differenc
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The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,
Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
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Considering the Ohm's law, the resistance of the lightbulb is 144.58 Ω.
<h3>Definition of current</h3>The flow of electricity through an object, such as a wire, is known as current (I). Its unit of measure is amps (A). So the current is a measure of the speed at which the charge passes a given reference point in a specified direction.
<h3>Definition of voltage</h3>The driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V) (voltage can also be referred to as the potential difference or electromotive force). That is, voltage is a measure of the work required to move a charge from one point to another.
<h3>Definition of resistance</h3>Resistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).
<h3>Ohm's law</h3>Ohm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.
This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents.
Mathematically, Ohm's law is expressed as:
Where:
- I is the current measured in amps (A).
- V the voltage measured in volts (V).
- R the resistance that is measured in ohms (Ω).
In this case, you know that the voltage between two points in a circuit is 120 V and there is a current of 0.83 A.
Replacing in the Ohm's Law:
Solving:
0.83 A× R= 120 V
<u><em>R= 144.58 Ω</em></u>
Finally, the resistance of the lightbulb is 144.58 Ω.
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