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steposvetlana [31]
3 years ago
15

A 9.0 cm object is 3.0 cm from a lens, which has a focal length of –12.0 cm.

Physics
2 answers:
artcher [175]3 years ago
7 0
The answers are -2.4 for the distance. 7.2 for the height. Concave for the type
g100num [7]3 years ago
3 0

Answer :

Image distance = -2.4 cm

Height of image = 7.2 cm

Concave lens

Explanation :

Given that,

Height of the object, h = 9 cm

Object distance, u = 3 cm

Focal length, f = -12 cm

Using mirror's formula

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

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

\dfrac{1}{v}=\dfrac{1}{(-12)}-\dfrac{1}{(3)}

v = -2.4 cm

Image distance is 2.4 cm from the lens.

Magnification, m =\dfrac{-v}{u}=\dfrac{h'}{h}

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

h' = 7.2 cm

The height of the image is 7.2 cm. This shows that the image is upright.

The lens used is concave lens.

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y = vertical distance moved.

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P = (0 + 2750m, 0) = (2750m, 0).

"it lifts off at an angle of 37 degrees from the horizontal. The plane has traveled 1.8 km since its wheels left the ground."

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The distance between two points (a, b) and (c, d) is:

D = √( (a  c)^2 + (b - d)^2)

In this case the points are:

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(4.188km, 1.083km) for the final position.

And the displacement will be:

D = √( (4.188km - 0)^2 + (1.083 - 0)^2) = 4.236km

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