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allochka39001 [22]
3 years ago
12

What are the Units of voltage?

Physics
2 answers:
Luba_88 [7]3 years ago
7 0
Voltage describes the amount of energy associated with electric charge as it moves around in a circuit. Its standard unit, the volt, is related to the standard units of electric charge (the coulomb) and energy (the joule) such that one volt is equal to one joule of energy for each coulomb of electric charge. :)
NeTakaya3 years ago
7 0

Answer:

Actually volts is itself a SI Unit of Electric potential, electromotive force

Symbol V

Named after Alessandro Volta

In SI base units: kg·m2·s−3·A−1

The volt (symbol: V) is the derived unit for electric potential, electric potential difference (voltage), and electromotive force. The volt is named in honour of the Italian physicist Alessandro Volta (1745–1827), who invented the voltaic pile.

One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, It can also be expressed as amperes times ohms (current times resistance, Ohm's law), watts per ampere (power per unit current, Joule's law), or joules per coulomb (energy per unit charge), which is also equivalent to electron-volts per elementary charge:

The "conventional" volt, V90, defined in 1988 by the 18th General Conference on Weights and Measures and in use from 1990, is implemented using the Josephson effect for exact frequency-to-voltage conversion, combined with the caesium frequency standard. For the Josephson constant, KJ = 2e/h (where e is the elementary charge and h is the Planck constant), the "conventional" value KJ-90 is used:

I hope you understand now

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Answer:

3.14946 rad/s

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In this system the angular momentum is conserved

L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s

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3 0
3 years ago
When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.
olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

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