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Fed [463]
3 years ago
13

What happens if two small positively charged particles of equal magnitude are placed close to each other? A) The particles will

experience a repulsive force but remain stationary. B) The particles will experience an attractive force causing them to meet. C) The particles will experience a repulsive force that will drive them apart along the line joining their centers. D) The particles will experience a force that will cause them to move in a circular path of uniform motion with fixed distance of separation.
Physics
2 answers:
serious [3.7K]3 years ago
7 0

That is not correct, it is C :)

elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

the correct answer is "c"

Explanation:

i took the test on usatestprep the answer is c (The particles will experience a repulsive force that will drive them apart along the line joining their centers.)

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Evan drew a diagram to illustrate radiation.
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D. Electromagnetic waves.

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B. 6

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Tyler drives 50km north. Tyler then drives back 30km south. What distance did he cover? What was his displacement?
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Answer:

he covered 80km his displacement was 20km

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Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen
Marta_Voda [28]

Answer:

a) that laser 1 has the first interference closer to the central maximum

c) Δy = 0.64 m

Explanation:

The interference phenomenon is described by the expression

         d sin θ = m λ

Where d is the separation of the slits, λ the wavelength and m an integer that indicates the order of interference

For the separation of the lines we use trigonometry

        tan θ = sin θ / cos θ = y / x

In interference experiments the angle is very small

          tan θ = sin θ = y / x

         d y / x = m λ

a) and b) We apply the equation to the first laser

          λ = d / 20

          d y / x = m d / 20

          y = m x / 20

          y = 1 4.80 / 20

          y = 0.24 m

The second laser

        λ = d / 15

          d y / x = m d / 15

          y = m x / 15

          y = 0.32 m

We can see that laser 1 has the first interference closer to the central maximum

c) laser 1

They ask us for the second maximum m = 2

            y₂ = 2 4.8 / 20

            y₂ = 0.48 m

For laser 2 they ask us for the third minimum m = 3

In this case to have a minimum we must add half wavelength

         y₃ = (m + ½) x / 15

         m = 3

         y₃ = (3 + ½) 4.8 / 15

         y₃ = 1.12 m

        Δy = 1.12 - 0.48

        Δy = 0.64 m

4 0
3 years ago
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