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JulsSmile [24]
3 years ago
14

E20E20: The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to

the virtual image you see of yourself in a plane mirror?
Physics
2 answers:
Lubov Fominskaja [6]3 years ago
5 0

The correct answer (sample response) is:

The image seems to be behind the mirror, but nothing is really there.

Include the following in your response:

The image appears to be behind the mirror.

If someone looks behind the mirror, there is no image there.

|Huntrw6|

Alika [10]3 years ago
3 0

Answer:

A plane mirror forms a virtual, erect, same size and laterally inverted image. Virtual means that the image is not real and cannot be obtained on screen. It means the rays appear to meet behind the mirror but when you will actually go behind, you will find nothing. A virtual image is always erect.

A real image is always inverted. It can be obtained on the screen. Rays actually meet at a point.

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Newton’s law of universal gravitation stars that every object in the universe attracts every other object
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Yes it does !  Uh huh. Right you are. Truer words are seldom written.

You have quoted the law quite accurately but also incompletely.  

Do you have a question to ask ?

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Water has physical properties that can be<br> measured and
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If you walk at an average speed of 5 km/h for 30 minutes, how
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Time=30min
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2 years ago
The law of mass action suggests that _____.
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the higher concentration of molecules, the faster a reaction can occur

7 0
3 years ago
Read 2 more answers
Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b
sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}

q=7.56\times10^{-7}C

Thus, the charge on each sphere is q=7.56\times10^{-7}C.

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}

q=3.78\times10^{-7}C

Thus, the charge on second sphere is q=3.78\times10^{-7}C and the charge on first sphere is 4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C.

6 0
3 years ago
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