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Alborosie
3 years ago
8

Which type of mirror produces images that are always upright and at the same distance from the mirror as the object is?

Physics
2 answers:
Nezavi [6.7K]3 years ago
8 0
<h3><u>Answer;</u></h3>

Flat mirror

<h3><u>Explanation;</u></h3>
  • <em><u>A flat mirror or a plane mirror forms images that are always at equal distance as the object from the mirror. The images formed by a pane mirror are always virtual and the size of the image is the same as that of the object.</u></em>
  • <em><u>Plane mirrors and convex mirrors only produces virtual images. Concave mirrors are the only mirrors that are capable of forming real images and occurs only when the object is located a distance greater than a focal length of the mirror.</u></em>
Alex73 [517]3 years ago
7 0
Well, that would be a plane (flat) mirror 
<span>provided that </span>
<span>the mirror and the object are oriented parallel to each other</span>
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Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

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E=mc^{2}

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Now the energy produced by the reactor in 1 year equals

Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules

Thus the mass that is covertred at 100% efficiency is

mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg

Part 2)

At 30% efficiency the mass converted equals

mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg

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3 years ago
A parallel-plate capacitor connected to a battery becomes fully charged. After the capacitor from the battery is disconnected, t
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Answer:

C.)The energy stored in the capacitor doubles its original value.

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C₂ = ∈A/2d.............................(2a)

From equations (1) and (2), the relationship between C₁ and C₂ is:

C₂ = C₁/2....................................(2b)

The original  Energy stored in the capacitor is given by :

E₁ = Q²/2C₁...............................(3)

On doubling the separation between the capacitance plates

E₂ = Q²/2C₂...............................(4a)

E₂ = Q²/2  *  1/C₂........................(4b)

Putting equation (2b) into (4b)

E₂ = Q²/2  *  2/C₁

E₂ = Q²/C₁.....................................(5)

Comparing equations (3) and (5)

E₂ = 2E₁

Therefore, the energy stored in the capacitor doubles its original value

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