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3 years ago

Hi

Physics

2 answers:

vaieri [72.5K]3 years ago

5 0

<h2>Explanation:</h2>

m = h₂/h₁

Here,

m = Magnification
h₁ = Height of the object
h₂ = Height of the image formed by the concave mirror

Magnification given is (-3). The linear magnification here is negative.

This implies that h₂ < h₁ because m < 1.

Therefore, the image formed is real and inverted. It is smaller than the object. So, the image is formed between the Centre of Curvature & the Focus.

[NOTE: If the image is highly reduced, it is formed at the focus.]

romanna [79]3 years ago

3 0

Answer:HERE IS YOUR ANSWER

THE POINTS ARE

IF WE PUT THE OBJECT BETWEEN THE FOCUS

AND THE POLE

THEN THE IMAGE FORMED WILL be MAGNIFIED

If that’s not what you are looking for, try this one:

For concave mirror the virtual image is formed when the object is kept in between the pole and the focus.

Given here the "size of the image" is twice to that of the object.

Hence, it is consider that the magnification is +2.

So, the magnification value is positive and the image formed will be "virtual and erect". Thus, the object should be kept in between the "pole and the focus" in concave mirror."

Explanation:

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Answers:

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b) time=1.6 s, frquency=0.625 Hz

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Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength $\lambda$ is the distance between to crests; this means we have $4\lambda$ in $40 m$ :

$40 m=4\lambda$

Clearing $\lambda$ :

$\lambda=\frac{40 m}{4}$

$\lambda=10 m$

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- $T$

Then:

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On the other hand, the frequency $f$ of the wave has an inverse relation with its period $T$ :

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$f=\frac{1}{1.6 s}$

$f=0.625 Hz$ This is the frequency of the wave

c) The speed $v$ of a wave is given by the following equation:

$v=\frac{\lambda}{T}$

$v=\frac{10 m}{1.6 s}$

Finally:

$v=6.25 m/s$

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