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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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A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil
snow_lady [41]

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1.137278672 m/s

+5.9 cm or -5.9 cm

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Maximum speed is given by

v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m

The locations are +5.9 cm or -5.9 cm

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