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Blizzard [7]
3 years ago
6

Frequency of the wave below?

Physics
1 answer:
agasfer [191]3 years ago
6 0

It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?

If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.

Frequency = speed/wavelength

Frequency = 2m/s /0.2444m

Frequency = 8.18 Hz

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alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

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Solution:

Magnification of lens, m = -\frac{v}{u}

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u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

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3 years ago
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