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zmey [24]
3 years ago
10

Hi

Physics
2 answers:
vaieri [72.5K]3 years ago
5 0
<h2><u>Explanation:</u></h2>

<em>m = h₂/h₁</em>

Here,

  • <em>m</em> = Magnification
  • <em>h₁ </em>= Height of the object
  • <em>h₂</em> = Height of the image formed by the concave mirror

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

This implies that <em>h₂ < h₁</em> because <em>m < 1</em>.

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

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

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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Answer:

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Explanation:

5 0
3 years ago
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Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3&gt;s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
3 years ago
Please help on this one
mixas84 [53]
Ep=mgh
h= Ep/mg
h=57÷(3.3×9.8)
h= 57÷32.34
h= 1.8m
So; the answer is B. 1.8m
6 0
3 years ago
Please help me someone !
san4es73 [151]

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

7 0
3 years ago
If you wanted the pitch of a horn to drop relative to an observer, which way would you move the horn, relative to where that obs
Vladimir [108]
We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.

This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.

When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength. 

Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.

From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>
3 0
3 years ago
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