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Artemon [7]
3 years ago
15

Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

.33m. A third charge of 1.0 nC is placed 1.9 m away from q1. If the net electric force on this third charge is zero, what is q2? (Give your answer in nC
Physics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

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A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2
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The acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

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Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far
olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

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θ = angle = ?

Using the equation

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Using the equation for small angles

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y = 0.0667 m

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3 years ago
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