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3 years ago

A 15.0 cm tall object is placed 20.0 cm from a converging lens with a 10.0 cm focal length. Where is the image located?

Physics

1 answer:

Ksju [112]3 years ago

4 0

Answer: 20 cm

Explanation:

Given that

F = 10 cm

U = 20 cm

1/F = 1/V - 1/U

1/10 = 1/V - 1/20

1/V = 1/10 - 1/20

1/V = (2 - 1)/20

1/V = 1/20

V = 20 cm

The image will be located at 20cm away from the lens.

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The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

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$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

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