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denis23 [38]
1 year ago
7

What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

electric field of magnitude 590 N/C
Physics
1 answer:
Pani-rosa [81]1 year ago
8 0

     charge must be equal to 5.74 ×10⁻⁵

 In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

     →    Fnet =0

     →    mg =  qE

 substituting the values we get :

         0.00345 × 9.81 =  q × 590

   →       q = 5.74 ×10⁻⁵

    Hence the charge must be equal to   5.74 ×10⁻⁵.

   Learn more about charges here:

          brainly.com/question/26092261

                    # SPJ4

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please mark this answer as brainlist

5 0
2 years ago
It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w
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Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

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As we know the formula:

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And we also know that :

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F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0
3 years ago
A flashlight converts?
sammy [17]

Answer:

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Explanation:

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Light energy is also called electromagnetic energy.

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4 0
2 years ago
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Explanation:

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3 0
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