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Aleksandr [31]
3 years ago
8

A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl

e moves from position A to position B. The electric potential at A is -55.0 V, and the electric potential at B is +27.0 V. Determine the charge of the particle. Include the algebraic sign ( + or - ) with your answer.
Physics
1 answer:
Sati [7]3 years ago
6 0

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

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Answer:1.55 times

Explanation:

Given

First wavelength(\lambda _1)=450 nm

Second wavelength(\lambda _2)=700 nm

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\lambda T=constant

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Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

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18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

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