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yarga [219]
3 years ago
8

PLEASE HELP Please explain to me what longitudinal and transverse wave are!

Physics
1 answer:
german3 years ago
6 0

Answer:

In a transverse wave the particles of the medium move perpendicular to the wave's direction of travel.

In a longitudinal wave, the particles of the medium move parallel to the wave's direction of travel.

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A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im
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Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

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We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}

\dfrac{1}{v}=\dfrac{7}{690}

v=98.57\ cm

We need to calculate the height of the image

Using formula of height

\dfrac{h'}{h}=\dfrac{-v}{u}

Where, h' = height of image

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h'=\dfrac{98.57\times4}{30}

h'=13.14\ cm

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}

\dfrac{1}{v}=-\dfrac{10}{299}

v=-29.9\ cm

We need to calculate the height of the image

\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}

h=-\dfrac{29.9\times4}{13.0}

h=-9.2\ cm

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Hence, This is required solution.

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3 years ago
A copper wire expand when
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