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Mariana [72]
3 years ago
12

Which of the following statements are true about the lenses used in eyeglasses to correct nearsightedness and farsightedness? (T

here may be more than one correct choice.)A: They produce a real image.B: They produce a virtual image.C: Both nearsightedness and farsightedness are corrected with a converging lens.D: Both nearsightedness and farsightedness are corrected with a diverging lens
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

A: They produce a real image.

Explanation:

The images formed on the retina of the eye for a normal visibility must always be real.

Only a real image can be physically projected on any physical object whereas the virtual images are visible due to reflections.

  • The nearsightedness is corrected with the help of a concave lens since it is the condition of the eye lens remaining thick and curved to converge the rays entering the eyes after a shorter distance which results in their image formation even before the retinal surface so to initially diverge them a bit so that they converge on the retinal surface and form the image there we use concave lens. Vice-versa of the above justification in  the case of farsightedness.
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A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated to
Soloha48 [4]

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

f_{beat} = |f_2-f_1|

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

f_{beat} = |f_2-f_1|

5Hz = |f_2-440Hz|

f_2 = 440 \pm 5

f_2 = 445Hz or 435Hz

B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

5 0
3 years ago
A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A
MakcuM [25]

Answer:

The magnitude and algebraic sign of q is 14\sqrt{2}\ \mu C

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}

The electric force at opposite corner

F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}

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F=\sqrt{F_{1}^2+F_{2}^2}

Put the value into the formula

F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}

The electric force at second corner

F_{2}=\dfrac{-kq^2}{2r^2}

The net force acting on either of the charges is zero.

So,  F=F'

\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}

\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}

q=14\sqrt{2}\ \mu C

Hence, The magnitude and algebraic sign of q is 14\sqrt{2}\ \mu C

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Answer:

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