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Oxana [17]
3 years ago
7

Which of the following sets of characteristics describes the image formed by a plane mirror?A. Virtual and invertedB. Real and u

prightC. Virtual and uprightD. Real and invertedE. All the previous statements can be correct

Physics
1 answer:
tiny-mole [99]3 years ago
4 0
<h2>Answer: Virtual and upright</h2><h2 />

A plane mirror is a highly polished flat surface with a very high capacity to reflect incident light.  

We can understand in a better way how this works with the figure attached:  

1. The incident rays coming from the real object reach the mirror and  

2.are reflected following the law of Reflection.  

3.The prolongation of those reflected rays converge at a point that does not coincide with the actual position of the object. At that point the virtual image of the object is formed.  

4.Then, the reflected divergent rays are captured by our eye converging on the retina.  

Now, the image is said to be virtual because it is a copy of the object that looks as if the object is behind the mirror and not in front of it or on the surface, but it is not really there. However, it can be seen when we focus it with our eyes.  

In addition, the image formed is:  

symmetrical, because apparently it is at the same distance from the mirror  

the same size as the object.  

upright, because it retains the same orientation as the object.

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It should be noted that this value can not be measured directly by experiments, but it can be determined indirectly by means of calculations from other atomic or molecular properties of the element. That is why the scale created by Pauling is an arbitrary scale, where the maximum value of electronegativity is 4, assigned to Fluorine (F) and the <u>lowest is 0.7, assigned to Francium (Fr).</u>

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
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