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makkiz [27]
3 years ago
6

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always pr

oduces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.
Physics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

     1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

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A trumpet player, using her ear, hears 5 beats per second when she plays a note on her trumpet and simultaneously sounds a 440 H
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<h2>Frequency of trumpet wire = 445 Hz</h2>

Explanation:

The given  frequency of tuning fork is 440 Hz . It produces 5 beats with trumpet wire . That means , the frequency of wire can be 440 ± 5

It can be either 445 or 435

Now the length of wire is increased , by which its frequency decreases . Because frequency is inversely proportional to length of wire .

If we decrease the frequency in 435 , the difference between tuning fork frequency and wire frequency will become greater than 5 even . So it cannot produce 3 beats with it .

If we decrease frequency from 445 , it can become 443 Hz . It gives 3 beat with the tuning fork as given .

Thus the initial frequency of wire is 445 Hz

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Answer:

a) v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s

b) v=+9.97m/s

Explanation:

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x_{1} =15m, t_{1}=3s

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v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }

a) For the three intervals:

v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s

v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s

v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s

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v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s

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