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34kurt
3 years ago
7

PLLLLLLLLLLLZ HELP

Physics
1 answer:
Katyanochek1 [597]3 years ago
6 0

The second choice is the correct one.

You can spot it just from the units. An equation can't be true unless it has the same units on both sides.

Wavelength has units of (meters).

Frequency has units of (per second).

When you multiply them, you get (meters per second), the unit of speed, and the 'c' on the other side is speed.

None of the other choices has units of speed on the right side.

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A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
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A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

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Explanation:

Hope this will be help!

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