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Masteriza [31]
3 years ago
8

Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as

originating from a point on the near side of the mirror.
Physics
1 answer:
Reil [10]3 years ago
5 0

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

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It can go back to it's original shape

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1) For a positive point charge, the lines radiate ………. . While, for a negative point charge, the lines converge …………. .
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Explanation:

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Hope dis help

5 0
3 years ago
A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.
Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

\epsilon=NA\dfrac{dB}{dt}

For N = 1

\epsilon=A\dfrac{dB}{dt}

We need to calculate the current

Using formula of current

i=\dfrac{\epsilon}{R}

Put the value of emf

i=\dfrac{A\dfrac{dB}{dt}}{R}

Now, if the number of turn is 22 , then induced emf would be

\epsilon'=NA\dfrac{dB}{dt}

Then the current would be

i'=\dfrac{\epsilon'}{NR}

i'=\dfrac{NA\dfrac{dB}{dt}}{NR}

i'=\dfrac{A\dfrac{dB}{dt}}{R}

i'=i

Hence, The current would be same in both situation.

4 0
3 years ago
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