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Alex777 [14]
3 years ago
6

A convex mirror is placed to the right of an object. The image formed by the mirror will be a

Physics
2 answers:
Harrizon [31]3 years ago
6 0

I think it’s C but I’m not too sure

Amanda [17]3 years ago
4 0
<h3><u>Answer;</u></h3>

C.) Virtual Image that appears on the right side of the mirror.

A convex mirror is placed to the right of an object. The image formed by the mirror will be a  <u><em>Virtual Image that appears on the right side of the mirror. </em></u>

<h3><u> Explanation;</u></h3>
  • <em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
  • A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.



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3 years ago
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4 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
- farmer has recently bought a 40 acre land and wishes to fence the perimeter. given that the Land is square, what lenght of fen
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Answer:

Explanation:

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Answer: a. Mass per unit length =0.0245kg/m

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C. Tension = 2.45 x10^-8N

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Explanation:

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3 years ago
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