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olchik [2.2K]
2 years ago
15

The special relationship between the ohm, the ampere, and the volt can be expressed using what?​

Physics
1 answer:
german2 years ago
8 0

Answer:

Ohm's law: I = V / R

Explanation:

V, I, and R, the parameters of Ohm's law

with

V = voltage in Volt

I = current in Ampere

R = resistance in Ohm

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. Introducing the constant of proportionality.

The following equation is known as Ohm's law:

I = V / R

You can also write Ohm's law as

V = I * R

You can also write Ohm's law as

R = V / I

Ohm's law states that the R in this relation is constant, independent of the current

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What is the energy of a photon whose frequency is 6.0 x 10^20?
omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

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