Physics, 11.10.2020 01:01 dacey4164

Two point charges are placed over the x-axis as follows: charge q1=+3nC is at x=0.26m, and charge q2=+6nC is at x=-0.45m. Find the magnitude and direction of the total force exerted by these two charges over a negative point charge q3=-8nC that is at x=0m.

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Answer from: sameh0425

F = 5.33 10⁻⁶ N , directed to the left

Explanation:

For this exercise we will use the force addition law where each force is electric given by Coulomb's law

F = ∑ F

F = F₁₃ + F₂₃

where the force is given by

F = k q₁ q₂ / r₁₂²

in our case

F₁₃ = k q₁ q₃ / r₁₃²

F₂₃ = k q₂ q₃ / r₂₃²

F = k q₃ (q₁ / r₁₃² + q₂ / r₂₃²)

as the charges are of the same sign the electric force is repulsive and due to the position of the charges it is directed to the left

let's calculate

F = 9 10⁹ 8 10⁻⁹ (3 10⁻⁹ / (0.26 - 0) 2 + 6 10⁻⁹ / (0.45-0) 2)

F = 72 (44,379 + 29,630) 10⁻⁹

F = 5.33 10⁻⁶ N

directed to the left

Answer from: Quest

nec chapter 5 contains

c. information on special occupancy areas

nec chapter 6 covers

a. special equipment

which of the following rules contain the words shall or shall not and must be adhered to?

b. mandatory rules

which of the following rules identify actions that are allowed but not required and typically cover options or alternative methods?

c. optional rules

which of the following serves as the introduction to the nec and provides an overview of the book?

a. nec article 90

organizations that test products for safety are

c. nrtls

the nec is published by the

a national fire protection association

the nec is divided into

b. 9 chapters

the nec provides informative annexes for

c. informational materials and examples

two terms from nec article 100 that trainees should become familiar with are

b. labeled and listed

explanation:

Answer from: Quest

answer: the correct answer id option(d).

explanation:

snell's law refraction:

$\frac{n_2}{n_1}=\frac{sin \theta_1}{sin\theta_2}$

the angle of refraction = $\theta_2=38 ^o$

the angle of incidence = $\theta_1=?$

refractive index of air = $n_1$ = 1

refractive index of glycerol $n_2$ = 1.47

$\frac{1.47}{1}=\frac{sin\theta_1}{sin 38^o}, (sin38^o=0.61)$

$sin\theta_2=0.8967$

$\theta_2=24.2^o, (sin^{-1}0.8967=63.72^o)$

from the given options, the most closest answer to 63.72° is 65° of option (d).

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Two point charges are placed over the x-axis as follows: charge q1=+3nC is at x=0.26m, and charge q2...

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