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Serjik [45]
3 years ago
14

Which statements describe the image produced by a concave lens?check all that apply.

Physics
2 answers:
ikadub [295]3 years ago
8 0

Answer:

A,E

Explanation:

A-the image is right-side up

B-the image is virtual

kotegsom [21]3 years ago
7 0

<em>Answer:</em>

<em>The answers are: </em>

  • <em>A-which is the image is always right side up.</em>
  • <em>E-the image is virtual</em>

<em></em>

<em>Explanation: MY EXPLANATION IS YOU ARE WELCOME BIG DOG 100..</em>

<em></em>

<h2 />
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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $\frac{2\pi r}{4} = $\frac{\pi R}{2}

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $\frac{\pi R}{2v} = \frac{\pi R}{40}

Hence, the time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4



 



6 0
1 year ago
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